Question

The following three substances are placed together in thermal contact:

1) Mg = 1.0 kg gold at temperature Tg = −13 C

2) Ms = 2.0 kg silver at temperature Ts = +7.0 C

3) Mi = 0.01 kg ice at temperature Ti = 0.0 C

What is the final equilibrium temperature T ?

Data:

specific heat of ice: Ci = 2100

specific latent heat of fusion of ice/water: Lf = 3.35 × 10^5

specific heat of gold: Cg = 126

specific heat of silver: Cs = 234

Answer 1

Heat required for gold to change its temperature from $T_g=-13\degree C$ to  $T=0\degree C$  is  $H_g=M_gC_g(T-T_g)=1.0*126*(0+13)=1638\,J$

Heat given out by silver to change its temperature from $T_s=7\degree C$ to  $T=0\degree C$  is  $H_s=M_sC_s(T_s-T)=2.0*234*(7-0)=3276\,J$

Heat required for ice at $T_i=0\degree C$ to change in to water at $T=0\degree C$ is

$H_i=m_iL=0.01*3.35*10^5=3350\,J$

Out of the heat $H_s=3276\,J$  given by silver, a part of it $1638\,J$ is used to raise the temperature of gold from $-13\degree C$ to $0\degree C$ and remaining heat $3276-1638=1638\,J$ is used by a part of ice to change its state into water at $0\degree C$.

A part of ice changes to water and the remaining will be left as ice $0\degree C$.

Hence the equilibrium temperature of the system will be $0\degree C$.