Question 10.5: A metal pellet with a mass of 100.0 g, originally at 88.4°C,...
A metal pellet with a mass of 100.0 \text{g}, originally at 88.4°\text{C}, is dropped into 125 \text{g} of water originally at 25.1°\text{C}. The final temperature of both the pellet and the water is 31.3°\text{C}. Calculate the heat capacity C (in \text{J/°C}) of the pellet.
Strategy Water constitutes the surroundings; the pellet is the system. Use Equation 10.14 (q_\text{surr} = sm∆T) to determine the heat absorbed by the water; then use Equation 10.12 (q = C∆T) to determine the heat capacity of the metal pellet.
Setup m_\text{water} = 125 \text{g}, s_\text{water} = 4.184 \text{J/g ⋅ °C}, and ∆T_\text{water} = 31.3°\text{C} – 25.1°\text{C} = 6.2°\text{C}. The heat absorbed by the water must be released by the pellet: q_\text{water} = – q_\text{pellet}, m_\text{pellet} = 100.0 \text{g}, and ∆T_\text{pellet} = 31.3°\text{C} – 88.4°\text{C} = –57.1°\text{C}.
From Equation 10.14, we have
q_\text{water} = \frac{4.184 \text{J}}{\text{g ⋅ °C}} × 125 \text{g} × 6.2°\text{C} = 3242.6 \text{J}
Thus,
q_\text{pellet} = –3242.6 \text{J}
From Equation 10.12, we have
–3242.6 \text{J} = C_\text{pellet} × (–57.1°\text{C})
Thus,
C_\text{pellet} = 57 \text{J/°C}