An 80 g aluminum calorimeter contains 350 g of water at an equilibrium temperature of 2 0 ∘ C . A 100 gm piece of metal, initially at 34 7 ∘ C is added to the calorimeter. The final temperature at equ

Question

An 80 g aluminum calorimeter contains 350 g of water at an equilibrium temperature of 2 0 ∘ C . A 100 gm piece of metal, initially at 34 7 ∘ C is added to the calorimeter. The final temperature at equilibrium is 3 2 ∘ C . Assume there is no external heat exchange. The specific heat capacities of aluminum and water are 910 and 4190 J/kg .K, respectively. The specific heat capacity of the metal is closest to:

TestHub · Accepted Answer

Heat lost by the metal = heat gained by the aluminum and the water:
$m_{m} s_{m} (347 - 32) = m_{A l} s_{A l} (32 - 20) + m_{w} s_{w} (32 - 20)$
$0.1 \times s_{m} \times 315 = 0.08 \times 910 \times 12 + 0.35 \times 4190 \times 12$
$31.5 s_{m} = 873.6 + 17598$
$s_{m} = \frac{18471.6}{31.5} \approx 586.4 J/kg.K$

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