Combining: \( T_\text{F} = \frac{9}{5}(T_\text{K} - 273.15) + 32 \) and \( T_\text{K} = \frac{5}{9}(T_\text{F} - 32) + 273.15 \). For temperature differences, \( 1\,\text{K} = 1\,°\text{C} \), and \( \Delta T_\text{F} = \frac{9}{5}\,\Delta T_\text{C} \).
Examples: 0°C = 273.15 K = 32°F. 100°C = 373.15 K = 212°F. 25°C = 298.15 K = 77°F. Body temperature 37°C = 310.15 K ≈ 98.6°F.
1. Temperature and temperature difference
Temperature \( T \) (in °C or K) measures how hot or cold a body is. The temperature difference \( \Delta T \) between two states is the change in temperature when the body is heated or cooled:
\[ \Delta T = T_\text{final} - T_\text{initial} \]
When we heat a body (add thermal energy), its temperature usually increases, so \( \Delta T > 0 \). When it cools (loses thermal energy), \( \Delta T < 0 \). In formulas we often use \( |\Delta T| \) for the magnitude, or define \( \Delta T \) so that the heat added is positive. Units: degrees Celsius (°C) or kelvin (K); for differences, \( 1\,\text{K} = 1\,°\text{C} \).
Example: Water is heated from 20°C to 80°C. Then \( \Delta T = 80 - 20 = 60\,\text{K} \) (or 60°C). If it cools from 80°C to 20°C, \( \Delta T = 20 - 80 = -60\,\text{K} \); the change is −60 K (temperature decreased).
2. Heat and specific heat capacity
Heat \( Q \) is energy transferred because of a temperature difference. When we say “heat supplied” or “energy transferred by heating”, we mean \( Q \) (in joules, J). The amount of heat needed to change the temperature of a mass \( m \) by \( \Delta T \) depends on the material:
\[ Q = m c \, \Delta T \]
where \( c \) is the specific heat capacity (or specific heat) of the material. It is the energy needed to raise the temperature of 1 kg of the material by 1 K (or 1°C). Units: \( c \) in \( \text{J/(kg}\cdot\text{K)} \) or \( \text{J/(kg}\cdot°\text{C)} \).
So: same mass and same \( \Delta T \), a larger \( c \) means more heat is needed. Water has a high specific heat capacity (\( c \approx 4200\,\text{J/(kg}\cdot\text{K)} \)); metals have lower \( c \) (e.g. copper \( \approx 390 \), aluminium \( \approx 900 \)).
Rearranging: \( c = Q/(m\,\Delta T) \); \( \Delta T = Q/(mc) \); \( m = Q/(c\,\Delta T) \). When the body loses heat (cools), \( Q < 0 \) if we take \( Q \) as “heat added to the body”; then \( \Delta T < 0 \). Alternatively, define \( Q \) as heat lost: \( Q_\text{lost} = mc|\Delta T| \).
Example: Heating 2 kg of water (\( c = 4200\,\text{J/(kg}\cdot\text{K)} \)) by 10 K requires \( Q = mc\Delta T = 2 \times 4200 \times 10 = 84\,000\,\text{J} = 84\,\text{kJ} \).
3. Energy change and conservation
When we heat a body and it does not do work or change phase, the increase in thermal energy (internal energy) of the body equals the heat supplied: \( \Delta U = Q \). So the energy “stored” as higher temperature is \( Q = mc\Delta T \).
Energy conservation: In an isolated system, total energy is constant. So: heat gained by one part = heat lost by another (if no other energy transfers). For two bodies in thermal contact reaching the same final temperature: \( Q_\text{gained} + Q_\text{lost} = 0 \), i.e. \( m_1 c_1 (T_f - T_1) + m_2 c_2 (T_f - T_2) = 0 \). Solve for final temperature \( T_f \).
\[ \text{Heat lost by hot body} = \text{Heat gained by cold body} \quad \Rightarrow \quad m_1 c_1 (T_1 - T_f) = m_2 c_2 (T_f - T_2) \]
Example: 0.5 kg water at 80°C is mixed with 1 kg water at 20°C (same \( c \)). Heat lost by hot = heat gained by cold: \( 0.5 \times c \times (80 - T_f) = 1 \times c \times (T_f - 20) \). So \( 40 - 0.5 T_f = T_f - 20 \), \( 60 = 1.5 T_f \), \( T_f = 40\,°\text{C} \).
4. Energy loss to the surroundings
In real situations, some heat is lost to the surroundings (air, container). So not all the energy supplied goes into raising the temperature of the body. We can write:
If we assume no losses, \( Q_\text{supplied} = mc\Delta T \). If we know that a fraction is lost, then \( Q_\text{to body} = \eta \, Q_\text{supplied} \) (e.g. \( \eta = 0.8 \) for 80% useful heating), so \( mc\Delta T = \eta \, Q_\text{supplied} \). Alternatively, “energy loss” might be given directly: \( Q_\text{lost} = Q_\text{supplied} - mc\Delta T \).
Example: A heater supplies 50 kJ. Only 40 kJ goes to heating 2 kg water (\( c = 4200 \)). Then \( 40\,000 = 2 \times 4200 \times \Delta T \), so \( \Delta T = 40\,000/8400 \approx 4.76\,\text{K} \). The remaining 10 kJ is “lost” (e.g. to the container or air).
5. Energy transformation: work and kinetic energy to heat
Energy can be transformed from one form to another. When friction does work, work → heat: the loss of mechanical energy appears as thermal energy. When a moving object is brought to rest by friction, its kinetic energy is converted to heat:
\[ \text{KE lost} = \frac{1}{2}mv^2 \quad \Rightarrow \quad Q = \frac{1}{2}mv^2 \quad \text{(if all KE becomes heat)} \]
So if a mass \( m \) moving at speed \( v \) is stopped and all its KE is converted to heating the object (or the surface), the heat produced is \( Q = \frac{1}{2}mv^2 \). This heat can then raise the temperature of a body: \( Q = mc\Delta T \), so \( \Delta T = Q/(mc) = (mv^2/2)/(mc) = v^2/(2c) \) for the body that absorbed the energy.
Example: A 2 kg block sliding at 3 m/s is stopped by friction. KE lost = \( \frac{1}{2} \times 2 \times 9 = 9\,\text{J} \). If all 9 J goes into heating the block (\( c = 500\,\text{J/(kg}\cdot\text{K)} \)), then \( 9 = 2 \times 500 \times \Delta T \), so \( \Delta T = 9/1000 = 0.009\,\text{K} \) (very small).
6. Heating and kinetic energy combined
Some problems combine thermal energy and kinetic energy. For example: a hot object is dropped; it has initial KE (or gains KE as it falls) and thermal energy; when it hits the ground, some KE may convert to heat. Or: total energy = KE + thermal (e.g. \( E = \frac{1}{2}mv^2 + mc\Delta T \) relative to a reference).
Principle: Conserve total energy. If an object falls and hits the ground, \( \text{PE lost} = \text{KE gained} + \text{heat to surroundings} \), or \( \text{KE at impact} = \text{work done against friction} = Q \). If we heat a body and it also gains speed (e.g. in a closed system), \( Q_\text{supplied} = \Delta U + \Delta(\text{KE}) + \text{work} \) as appropriate.
Example: A 1 kg metal ball (\( c = 400\,\text{J/(kg}\cdot\text{K)} \)) at 100°C is dropped from 10 m. It hits the ground at speed \( v = \sqrt{2gh} \approx 14\,\text{m/s} \). KE at impact \( = \frac{1}{2} \times 1 \times 14^2 \approx 98\,\text{J} \). If 50% of this KE is converted to heating the ball, \( Q = 49\,\text{J} \), so \( \Delta T = 49/(1 \times 400) \approx 0.12\,\text{K} \). The ball’s temperature rises slightly on impact.
3. A heater supplies 126 kJ to 5 kg of water. If there is no energy loss, what is the rise in temperature? (\( c = 4200\,\text{J/(kg}\cdot\text{K)} \).)
Solution: \( Q = mc\Delta T \Rightarrow \Delta T = Q/(mc) = 126\,000/(5 \times 4200) = 126\,000/21\,000 = 6\,\text{K} \). So the temperature rises by 6°C (or 6 K).
4. 0.4 kg of water at 90°C is mixed with 0.6 kg of water at 20°C. Find the final temperature (same \( c \), no heat loss).
Solution: Heat lost by hot = heat gained by cold: \( 0.4 \times c \times (90 - T_f) = 0.6 \times c \times (T_f - 20) \). Cancel \( c \): \( 36 - 0.4 T_f = 0.6 T_f - 12 \), \( 48 = T_f \), so \( T_f = 48\,°\text{C} \).
5. An electric heater supplies 200 kJ. Due to energy loss, only 150 kJ goes to heating 10 kg of water. What is the temperature rise of the water? (\( c = 4200\,\text{J/(kg}\cdot\text{K)} \).)
Solution: Only \( Q = 150\,\text{kJ} = 150\,000\,\text{J} \) is used to heat the water. \( \Delta T = Q/(mc) = 150\,000/(10 \times 4200) = 150\,000/42\,000 \approx 3.57\,\text{K} \). The remaining 50 kJ is lost to the surroundings.
6. A 5 kg mass moving at 4 m/s is brought to rest by friction. If all the kinetic energy is converted to heat in the mass itself (\( c = 500\,\text{J/(kg}\cdot\text{K)} \)), find the temperature rise.
Solution: KE lost = \( \frac{1}{2}mv^2 = \frac{1}{2} \times 5 \times 16 = 40\,\text{J} \). This becomes heat: \( Q = 40\,\text{J} = mc\Delta T \), so \( \Delta T = 40/(5 \times 500) = 40/2500 = 0.016\,\text{K} \). Very small rise because the KE is small.
7. A 2 kg copper block (\( c = 390\,\text{J/(kg}\cdot\text{K)} \)) at 150°C is dropped into 5 kg of water at 20°C. Find the final temperature. (No heat loss; water \( c = 4200\,\text{J/(kg}\cdot\text{K)} \).)
Solution: Heat lost by copper = heat gained by water: \( m_c c_c (150 - T_f) = m_w c_w (T_f - 20) \). \( 2 \times 390 \times (150 - T_f) = 5 \times 4200 \times (T_f - 20) \). \( 780(150 - T_f) = 21\,000(T_f - 20) \), \( 117\,000 - 780 T_f = 21\,000 T_f - 420\,000 \), \( 537\,000 = 21\,780 T_f \), \( T_f \approx 24.7\,°\text{C} \). The water has much larger thermal capacity (\( m_w c_w \)), so the final temperature is close to the initial water temperature.
8. A 0.5 kg metal ball (\( c = 400\,\text{J/(kg}\cdot\text{K)} \)) at 80°C falls from rest through 20 m and hits the ground. If 30% of its kinetic energy at impact is converted to heating the ball, find the temperature rise of the ball. (Use \( g = 10\,\text{m/s}^2 \).)
Solution: Speed at impact: \( v^2 = 2gh = 2 \times 10 \times 20 = 400 \), so \( v = 20\,\text{m/s} \). KE = \( \frac{1}{2} \times 0.5 \times 400 = 100\,\text{J} \). 30% to heat: \( Q = 0.30 \times 100 = 30\,\text{J} \). \( \Delta T = Q/(mc) = 30/(0.5 \times 400) = 30/200 = 0.15\,\text{K} \). So the ball’s temperature rises by 0.15 K (or 0.15°C) on impact.
9. To raise the temperature of 4 kg of a substance by 15 K, 72 kJ of heat is needed. Find the specific heat capacity of the substance.
Solution: \( Q = mc\Delta T \Rightarrow c = Q/(m\Delta T) = 72\,000/(4 \times 15) = 72\,000/60 = 1200\,\text{J/(kg}\cdot\text{K)} \).
10. A 1 kg block at 25°C receives 50 kJ from a heater. If the block’s specific heat capacity is 500 J/(kg·K), what is its final temperature? (Assume no heat loss.)
Solution: \( Q = mc\Delta T \Rightarrow \Delta T = Q/(mc) = 50\,000/(1 \times 500) = 100\,\text{K} \). Final temperature \( T_f = T_\text{initial} + \Delta T = 25 + 100 = 125\,°\text{C} \).
Simple (11–30)
11. Water is heated from 15°C to 95°C. What is the temperature difference \( \Delta T \) in kelvins?
20. A 3 kg mass moving at 2 m/s is stopped by friction. If all its kinetic energy becomes heat in the mass (\( c = 600\,\text{J/(kg}\cdot\text{K)} \)), find the temperature rise.
Solution: KE \( = \frac{1}{2}mv^2 = \frac{1}{2} \times 3 \times 4 = 6\,\text{J} \). \( Q = 6 = mc\Delta T \), so \( \Delta T = 6/(3 \times 600) = 1/300 \approx 0.0033\,\text{K} \).
21. 2 kg of water at 80°C is mixed with 3 kg of water at 25°C. Find the final temperature. (No heat loss.)
26. Equal masses of water and oil are heated by the same heater for the same time. The water temperature rises by 5 K. If water has \( c = 4200 \) and oil has \( c = 2000\,\text{J/(kg}\cdot\text{K)} \), by how much does the oil temperature rise?
30. Express 50 kJ in joules. How much is that in heat needed to raise 1 kg of water by 1 K? (Use \( c = 4200\,\text{J/(kg}\cdot\text{K)} \).)
Solution: 50 kJ = 50\,000 J. To raise 1 kg water by 1 K we need \( Q = 1 \times 4200 \times 1 = 4200\,\text{J} \). So 50\,000 J raises 1 kg water by \( 50\,000/4200 \approx 11.9\,\text{K} \).
Medium (31–50)
31. A 1.5 kg iron block (\( c = 450\,\text{J/(kg}\cdot\text{K)} \)) at 200°C is dropped into 4 kg of water at 25°C. Find the final temperature. (No heat loss; water \( c = 4200\,\text{J/(kg}\cdot\text{K)} \).)
33. A 2 kg block at 6 m/s is stopped by friction. 40% of the kinetic energy heats the block (\( c = 500\,\text{J/(kg}\cdot\text{K)} \)). Find the temperature rise.
34. 0.5 kg water at 90°C, 0.5 kg water at 30°C, and 0.5 kg water at 50°C are mixed. Find the final temperature. (No heat loss.)
Solution: Total heat lost by hot = heat gained by cold. \( 0.5c(90 - T_f) + 0.5c(50 - T_f) = 0.5c(T_f - 30) \) (if \( T_f \) between 30 and 50). So \( 90 - T_f + 50 - T_f = T_f - 30 \), \( 170 = 3T_f \), \( T_f \approx 56.7\,°\text{C} \). Check: 90 and 50 lose heat, 30 gains; all consistent.
35. A 0.3 kg metal sphere (\( c = 380\,\text{J/(kg}\cdot\text{K)} \)) at 150°C is placed in 1.2 kg of water at 22°C. The final temperature is 28°C. Find the specific heat capacity of the metal (if we assumed the wrong value 380). Actually: verify that with \( c = 380 \) the final temperature would be about 28°C.
Solution: Heat lost by metal = heat gained by water: \( 0.3 \times c \times (150 - 28) = 1.2 \times 4200 \times (28 - 22) \). \( 0.3 \times c \times 122 = 30\,240 \), \( c = 30\,240/36.6 \approx 826\,\text{J/(kg}\cdot\text{K)} \). (So the metal’s \( c \) is about 826, not 380. If we use 380 we get \( T_f \) different from 28°C.)
36. An electric kettle supplies 200 kJ. 20 kJ is lost to the surroundings. The rest heats 2 kg of water from 20°C. What is the final temperature of the water?
37. A 1 kg ball (\( c = 400\,\text{J/(kg}\cdot\text{K)} \)) is dropped from 15 m. If 25% of its kinetic energy at impact is converted to heating the ball, find the temperature rise. (Use \( g = 10\,\text{m/s}^2 \).)
38. 1 kg of oil (\( c = 2100\,\text{J/(kg}\cdot\text{K)} \)) at 80°C is mixed with 2 kg of water at 25°C. Find the final temperature. (No heat loss; water \( c = 4200 \).)
39. A 5 kg block is heated from 15°C to 35°C. The heat supplied is 110 kJ. Find the specific heat capacity. If 10 kJ was lost to the surroundings, what would the effective \( c \) calculated from \( Q_\text{supplied} \) be?
Solution: No loss: \( c = Q/(m\Delta T) = 110\,000/(5 \times 20) = 1100\,\text{J/(kg}\cdot\text{K)} \). With 10 kJ lost, only 100 kJ heats the block: \( c_\text{eff} = 100\,000/(5 \times 20) = 1000\,\text{J/(kg}\cdot\text{K)} \) (underestimate if we use supplied heat).
40. A 3 kg copper block at 120°C is placed in 6 kg of water at 18°C. Find the final temperature. (Copper \( c = 390\,\text{J/(kg}\cdot\text{K)} \); water \( c = 4200 \).)
41. A car of mass 1200 kg moving at 20 m/s is braked to rest. If 50% of the kinetic energy is converted to heating the brake discs (total mass 8 kg, \( c = 500\,\text{J/(kg}\cdot\text{K)} \)), find the temperature rise of the discs.
Solution: KE \( = \frac{1}{2} \times 1200 \times 400 = 240\,000\,\text{J} \). \( Q = 0.5 \times 240\,000 = 120\,000\,\text{J} \). \( \Delta T = Q/(mc) = 120\,000/(8 \times 500) = 30\,\text{K} \). So the brake discs rise by 30°C.
42. 0.4 kg of water at 95°C is poured into 0.6 kg of water at 15°C in a container of negligible heat capacity. 5% of the heat is lost to the air. Estimate the final temperature (approximate, using average heat for the loss).
Solution: Without loss, heat balance: \( 0.4c(95 - T_f) = 0.6c(T_f - 15) \) gives \( T_f = 53\,°\text{C} \). With 5% loss, slightly less heat is available to warm the cold water; final \( T_f \) is a bit lower, roughly 51–52°C (exact needs iterative or effective reduced heat).
43. A 2 kg aluminium block (\( c = 900 \)) at 100°C and a 2 kg steel block (\( c = 500\,\text{J/(kg}\cdot\text{K)} \)) at 100°C are both dropped into 10 kg of water at 20°C. Which raises the water temperature more? By how much is the final temperature in each case? (No heat loss.)
Solution: Al: heat lost \( = 2 \times 900 \times (100 - T_f) \), water gain \( = 10 \times 4200 \times (T_f - 20) \). \( 1800(100 - T_f) = 42\,000(T_f - 20) \), \( 180\,000 = 43\,800 T_f - 840\,000 \), \( T_f \approx 23.3\,°\text{C} \). Steel: \( 2 \times 500 \times (100 - T_f) = 10 \times 4200 \times (T_f - 20) \), \( 1000(100 - T_f) = 42\,000(T_f - 20) \), \( T_f \approx 22.3\,°\text{C} \). Aluminium has higher \( c \), so it gives more heat and raises water more (23.3°C vs 22.3°C).
44. A heater supplies 60 kJ to 2 kg of water. The temperature rises by 6 K. Was there heat loss? If so, how much energy was lost?
Solution: Heat needed for \( \Delta T = 6\,\text{K} \) with no loss: \( Q_\text{need} = 2 \times 4200 \times 6 = 50\,400\,\text{J} = 50.4\,\text{kJ} \). Supplied 60 kJ, so lost \( = 60 - 50.4 = 9.6\,\text{kJ} \).
45. A 0.4 kg metal at 200°C is placed in 1.6 kg water at 25°C. The final temperature is 35°C. Find the specific heat capacity of the metal. (Water \( c = 4200\,\text{J/(kg}\cdot\text{K)} \).)
Solution: Heat lost by metal = heat gained by water: \( 0.4 \times c \times (200 - 35) = 1.6 \times 4200 \times (35 - 25) \). \( 0.4 \times c \times 165 = 67\,200 \), \( c = 67\,200/66 = 1018.\overline{18} \approx 1020\,\text{J/(kg}\cdot\text{K)} \).
46. 2 kg of water at 70°C is mixed with 3 kg of water at 25°C. What is the final temperature? If the mixture then loses 84 kJ to the surroundings, what is the new temperature?
47. A 1 kg object at 10 m/s is stopped; 60% of its KE heats the object (\( c = 400\,\text{J/(kg}\cdot\text{K)} \)) and 40% heats a 2 kg block (\( c = 500 \)) in contact. Find the temperature rise of each.
49. A 3 kg block is heated from 20°C to 80°C. The heater supplies 162 kJ. What percentage of the supplied energy was lost to the surroundings?
Solution: Heat needed (no loss): \( Q_\text{need} = mc\Delta T \). We need \( c \); if we assume \( c = 900 \) (e.g. aluminium), \( Q_\text{need} = 3 \times 900 \times 60 = 162\,000\,\text{J} \). So no loss. If we assume the block is water, \( Q_\text{need} = 3 \times 4200 \times 60 = 756\,000\,\text{J} \), so supplied 162 kJ would mean huge “loss” (actually the substance is not water). So the problem implies the block is like aluminium: 0% lost. Alternatively: if \( c \) is unknown, \( Q_\text{used} = 162\,\text{kJ} \) and \( \Delta T = 60\,\text{K} \), so \( mc = 162\,000/60 = 2700 \). Any loss would mean \( Q_\text{used} < 162\,\text{kJ} \). So for a metal with \( c \approx 900 \), 0% was lost.
50. A 1.2 kg copper bar (\( c = 390\,\text{J/(kg}\cdot\text{K)} \)) at 180°C is placed in 5 kg of water at 20°C. Find the final temperature. Then: how much heat was transferred from the copper to the water?