A copper calorimeter of mass 300 gm contains 500 gm of water at temperature [latex]15^o[/latex] C. A 560-gm block of aluminium of temperature [latex]100^oC[/latex] is dropped in the calorimeter and the temperature is observed to increase to [latex]22.5^o[/latex]C. Find the specific heat capacity of aluminium. (Specific heat capacity of copper [latex](s_c)[/latex] = [latex]400\ JKg^{-1}K^{-1}[/latex]).

Solution

Given,

Mass of calorimeter [latex](m_c)[/latex] = 300 gm = 0.3 kg

Mass of water [latex](m_w)[/latex] = 500 gm = 0.5 kg

Temperature [latex]\theta_1[/latex] = [latex]15^o[/latex]C

Mass of aluminium block [latex](m_a)[/latex] = 560 gm = 0.56 kg

Temperature [latex]\theta_2[/latex] = [latex]100^o[/latex]C

Temperature of mixture [latex]\theta[/latex] = [latex]22.5^o[/latex]C

Specific heat capacity of copper [latex](s_c)[/latex] = [latex]400 JKg^{-1}K^{-1}[/latex]

Specific heat capacity of aluminium [latex](s_a)[/latex] =?

According to principle of calorimetry,

Heat lost by aluminium = heat gained by calorimeter + heat gained by water

[latex]m_a s_a \Delta\theta = m_c s_c \Delta\theta + m_w s_w \Delta\theta[/latex]