Answer to Question #94485 in General Chemistry for Riley

Question

A 78.7 g piece of metal at 87.0°C is placed in 141 g of water at 21.0°C contained in a calorimeter. The metal and water come to the same temperature at 24.2°C. How much heat (in J) did the metal give up to the water? (Assume the specific heat of water is 4.18 J/g·°C across the temperature range.)

What is the specific heat (in J/g·°C) of the metal?

Answer 1

Heat given by metal = Heat absorbed by water =

$m_wc_w(T_2-T_1)=141\times 4.18\times(24.2-21)\\=1886\ J$

Heat released by metal = $m_mc_m(T_2-T_1)=1886\ J$

$1886=78.7\times c_m\times(78.7-24.2)\\c_m=\frac{1886}{78.7\times54.5}=0.44\ J\ g^{-1}\ \degree C^{-1}$

Specific heat of metal $=0.44\ J\ g^{-1}\ \degree C^{-1}$