A metal piece present at $120^{\circ} \mathrm{C}$ is quickly dropped in a calorimeter of mass 80 g containing 200 g of water at $30^{\circ} \mathrm{C}$. The final temperature attained by the mixture is $40^{\circ} \mathrm{C}$. Calculate the thermal capacity of the metal piece. [Specific heat capacity of water $=4.2 \mathrm{Jg}^{-1}{ }^{\circ} \mathrm{C}^{-1}$, Specific heat capacity of calorimeter $=0.4 \mathrm{jg}^{-1}{ }^{\circ} \mathrm{C}^{-1}$ ]
Explanation
To calculate the thermal capacity of the metal piece, we need to use the principle of heat transfer. The heat lost by the metal piece is equal to the heat gained by the water and the calorimeter. We can use the formula Q = mcΔT, where Q is the heat, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. By substituting the given values, we can calculate the thermal capacity of the metal piece.
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