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Difficulty: HardHeat Capacity and Specific Heat Capacity

A composite solid consists of a metal block AA of mass 2.0 kg2.0\text{ kg} and a metal block BB of mass 1.0 kg1.0\text{ kg} joined in thermal contact. Block AA has a specific heat capacity of 300 J kg−1K−1300\text{ J kg}^{-1}\text{K}^{-1}. When the composite solid absorbs 18 kJ18\text{ kJ} of thermal energy, its overall temperature increases by 15 K15\text{ K}. Assuming no heat loss to the surroundings, what is the specific heat capacity of metal block BB?

  1. 600 J kg−1K−1600\text{ J kg}^{-1}\text{K}^{-1}Answer
  2. B
    1200 J kg−1K−11200\text{ J kg}^{-1}\text{K}^{-1}
  3. C
    400 J kg−1K−1400\text{ J kg}^{-1}\text{K}^{-1}
  4. D
    200 J kg−1K−1200\text{ J kg}^{-1}\text{K}^{-1}

Answer

600 J kg−1K−1600\text{ J kg}^{-1}\text{K}^{-1}
The total heat energy supplied (18 kJ=18000 J18\text{ kJ} = 18000\text{ J}) causes a temperature rise of 15 K15\text{ K}, giving a total heat capacity Ctotal=1200 J K−1C_{\text{total}} = 1200\text{ J K}^{-1} for the composite solid. Subtracting block AA's heat capacity (CA=2.0 kg×300 J kg−1K−1=600 J K−1C_A = 2.0\text{ kg} \times 300\text{ J kg}^{-1}\text{K}^{-1} = 600\text{ J K}^{-1}) leaves a heat capacity of 600 J K−1600\text{ J K}^{-1} for block BB. Dividing block BB's heat capacity by its mass (1.0 kg1.0\text{ kg}) yields its specific heat capacity of 600 J kg−1K−1600\text{ J kg}^{-1}\text{K}^{-1}.

Step-by-Step Solution

1
Calculate the total heat capacity (CtotalC_{\text{total}}) of the composite system.
Ctotal=QΔT=18000 J15 K=1200 J K−1C_{\text{total}} = \frac{Q}{\Delta T} = \frac{18000\text{ J}}{15\text{ K}} = 1200\text{ J K}^{-1}
Heat capacity is defined as the total heat energy absorbed per unit change in temperature.
2
Determine the heat capacity (CAC_A) of metal block AA.
CA=mA×cA=2.0 kg×300 J kg−1K−1=600 J K−1C_A = m_A \times c_A = 2.0\text{ kg} \times 300\text{ J kg}^{-1}\text{K}^{-1} = 600\text{ J K}^{-1}
Heat capacity is the product of mass and specific heat capacity.
3
Determine the heat capacity (CBC_B) of metal block BB.
CB=Ctotal−CA=1200 J K−1−600 J K−1=600 J K−1C_B = C_{\text{total}} - C_A = 1200\text{ J K}^{-1} - 600\text{ J K}^{-1} = 600\text{ J K}^{-1}
Total heat capacity of a composite body is the sum of the individual heat capacities of its components.
4
Calculate the specific heat capacity (cBc_B) of metal block BB.
cB=CBmB=600 J K−11.0 kg=600 J kg−1K−1c_B = \frac{C_B}{m_B} = \frac{600\text{ J K}^{-1}}{1.0\text{ kg}} = 600\text{ J kg}^{-1}\text{K}^{-1}
Specific heat capacity is the heat capacity per unit mass of that specific substance.

Key Concept

Additivity of Heat Capacity in Composite Systems (Ctotal=mAcA+mBcBC_{\text{total}} = m_A c_A + m_B c_B)
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Topics

Heat Capacity and Specific Heat Capacity

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