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Specific Heat worksheet

Grades 9–10 · CCSS HSA.CED.A.4

Solve each problem and show your work. Round answers to 2 decimal places where needed.

1.
Problem 1.
- Solve for: Specific heat (c)
- Heat energy (Q): 850 kJ
- Mass (m): 2 kg
- Temperature change (ΔT): 64 °C
- Show specific heat in: Joules per kilogram kelvin (J/(kg·K))
Find: Specific heat
Answer
2.
Problem 2.
- Solve for: Heat (Q)
- Material: Steam (near 100 °C, constant pressure): ≈2,010 J/(kg·K) (typical)
- Mass (m): 2 kg
- Temperature change (ΔT): 55 °C
- Show heat in: Kilocalories (kcal)
Find: Heat energy
Answer
3.
Problem 3.
- Solve for: Temperature change (ΔT)
- Material: Dry air (constant pressure): ≈1,005 J/(kg·K) (typical)
- Heat energy (Q): 740 kJ
- Mass (m): 3 kg
- Show temperature change in: Degrees Celsius (°C)
Find: Temperature change
Answer
4.
Problem 4.
- Solve for: Heat (Q)
- Material: Olive oil: ≈1,970 J/(kg·K) (typical)
- Mass (m): 2 kg
- Temperature change (ΔT): 74 °C
- Show heat in: Kilojoules (kJ)
Find: Heat energy
Answer
5.
Problem 5.
- Solve for: Heat (Q)
- Material: Lead: ≈129 J/(kg·K) (typical)
- Mass (m): 3 kg
- Temperature change (ΔT): 90 °C
- Show heat in: British thermal units (IT) (BTU)
Find: Heat energy
Answer
6.
Problem 6.
- Solve for: Specific heat (c)
- Heat energy (Q): 720 kJ
- Mass (m): 3 kg
- Temperature change (ΔT): 80 °C
- Show specific heat in: BTU per pound degree Fahrenheit (BTU/(lb·°F))
Find: Specific heat
Answer
7.
Problem 7.
- Solve for: Temperature change (ΔT)
- Material: Ice (near 0 °C): ≈2,090 J/(kg·K) (typical)
- Heat energy (Q): 450 kJ
- Mass (m): 2 kg
- Show temperature change in: Degrees Fahrenheit (°F)
Find: Temperature change
Answer
8.
Problem 8.
- Solve for: Heat (Q)
- Material: Ethanol: ≈2,440 J/(kg·K) (typical)
- Mass (m): 1 kg
- Temperature change (ΔT): 65 °C
- Show heat in: Kilowatt-hours (kWh)
Find: Heat energy
Answer
9.
Problem 9.
- Solve for: Mass (m)
- Material: Ice (near 0 °C): ≈2,090 J/(kg·K) (typical)
- Heat energy (Q): 550 kJ
- Temperature change (ΔT): 66 °C
- Show mass in: Grams (g)
Find: Mass
Answer
10.
Problem 10.
- Solve for: Temperature change (ΔT)
- Material: Dry air (constant pressure): ≈1,005 J/(kg·K) (typical)
- Heat energy (Q): 700 kJ
- Mass (m): 2 kg
- Show temperature change in: Degrees Fahrenheit (°F)
Find: Temperature change
Answer

Answer key: Specific Heat worksheet

1. Specific heat: 6,640.63 J/(kg·K)
1. Change heat to J, the standard unit the formula uses (1 kJ = 1,000 J)
  $850 kJ \times 1,000$ $= 850,000 J$
2. To find the specific heat, divide the heat by the mass times the temperature change
  $\frac{850,000}{2 \times 64}$ $= 6,640.625 J/(kg·K)$
2. Heat energy: 52.84 kcal
1. Look up a typical specific heat (c) for steam (near 100 °C, constant pressure): the energy to warm 1 kg by 1 degree
  $c$ $= 2,010 J/(kg·K)$
2. Multiply the mass (m) by the specific heat (c) and the temperature change (ΔT) to get the heat taken in (Q)
  $2 \times 2,010 \times 55$ $= 221,100 J$
3. Change the answer to kcal, the unit you picked: divide by 4,184
  $\frac{221,100}{4,184}$ $= 52.844168 kcal$
3. Temperature change: 245.44 °C
1. Change heat to J, the standard unit the formula uses (1 kJ = 1,000 J)
  $740 kJ \times 1,000$ $= 740,000 J$
2. Look up a typical specific heat (c) for dry air (constant pressure): the energy to warm 1 kg by 1 degree
  $c$ $= 1,005 J/(kg·K)$
3. To find the temperature change, divide the heat by the mass times the specific heat
  $\frac{740,000}{3 \times 1,005}$ $= 245.439469 K$
4. Heat energy: 291.56 kJ
1. Look up a typical specific heat (c) for olive oil: the energy to warm 1 kg by 1 degree
  $c$ $= 1,970 J/(kg·K)$
2. Multiply the mass (m) by the specific heat (c) and the temperature change (ΔT) to get the heat taken in (Q)
  $2 \times 1,970 \times 74$ $= 291,560 J$
3. Change the answer to kJ, the unit you picked: divide by 1,000
  $\frac{291,560}{1,000}$ $= 291.56 kJ$
5. Heat energy: 33.01 BTU
1. Look up a typical specific heat (c) for lead: the energy to warm 1 kg by 1 degree
  $c$ $= 129 J/(kg·K)$
2. Multiply the mass (m) by the specific heat (c) and the temperature change (ΔT) to get the heat taken in (Q)
  $3 \times 129 \times 90$ $= 34,830 J$
3. Change the answer to BTU, the unit you picked: divide by 1,055.05585262
  $\frac{34,830}{1,055.05585262}$ $= 33.01247 BTU$
6. Specific heat: 0.72 BTU/(lb·°F)
1. Change heat to J, the standard unit the formula uses (1 kJ = 1,000 J)
  $720 kJ \times 1,000$ $= 720,000 J$
2. To find the specific heat, divide the heat by the mass times the temperature change
  $\frac{720,000}{3 \times 80}$ $= 3,000 J/(kg·K)$
3. Change the answer to BTU/(lb·°F), the unit you picked: divide by 4,186.799999999999
  $\frac{3,000}{4,186.799999999999}$ $= 0.716538 BTU/(lb·°F)$
7. Temperature change: 193.78 °F
1. Change heat to J, the standard unit the formula uses (1 kJ = 1,000 J)
  $450 kJ \times 1,000$ $= 450,000 J$
2. Look up a typical specific heat (c) for ice (near 0 °C): the energy to warm 1 kg by 1 degree
  $c$ $= 2,090 J/(kg·K)$
3. To find the temperature change, divide the heat by the mass times the specific heat
  $\frac{450,000}{2 \times 2,090}$ $= 107.655502 K$
4. Change the answer to °F, the unit you picked: divide by 0.555555555556
  $\frac{107.655502}{0.555555555556}$ $= 193.779904 °F$
8. Heat energy: 0.04 kWh
1. Look up a typical specific heat (c) for ethanol: the energy to warm 1 kg by 1 degree
  $c$ $= 2,440 J/(kg·K)$
2. Multiply the mass (m) by the specific heat (c) and the temperature change (ΔT) to get the heat taken in (Q)
  $1 \times 2,440 \times 65$ $= 158,600 J$
3. Change the answer to kWh, the unit you picked: divide by 3,600,000
  $\frac{158,600}{3,600,000}$ $= 0.044056 kWh$
9. Mass: 3,987.24 g
1. Change heat to J, the standard unit the formula uses (1 kJ = 1,000 J)
  $550 kJ \times 1,000$ $= 550,000 J$
2. Look up a typical specific heat (c) for ice (near 0 °C): the energy to warm 1 kg by 1 degree
  $c$ $= 2,090 J/(kg·K)$
3. To find the mass, divide the heat by the specific heat times the temperature change
  $\frac{550,000}{2,090 \times 66}$ $= 3.987241 kg$
4. Change the answer to g, the unit you picked: divide by 0.001
  $\frac{3.987241}{0.001}$ $= 3,987.240829 g$
10. Temperature change: 626.87 °F
1. Change heat to J, the standard unit the formula uses (1 kJ = 1,000 J)
  $700 kJ \times 1,000$ $= 700,000 J$
2. Look up a typical specific heat (c) for dry air (constant pressure): the energy to warm 1 kg by 1 degree
  $c$ $= 1,005 J/(kg·K)$
3. To find the temperature change, divide the heat by the mass times the specific heat
  $\frac{700,000}{2 \times 1,005}$ $= 348.258706 K$
4. Change the answer to °F, the unit you picked: divide by 0.555555555556
  $\frac{348.258706}{0.555555555556}$ $= 626.865672 °F$