Answer to Question #88908 in Molecular Physics | Thermodynamics for Sridhar

Question

one gram of ice at 0°C is mixed with one gram of water at 100°C the resulting temperature will be
1)5°C
2)0°C
3)10°C
4)100°C

Answer 1

$Q_1=Q_2+Q_3$

$Q_1$ - amount of heat transferred from the water cooling

$Q_2$ - amount of heat transferred for the ice melting

$Q_3$ - amount of heat transferred for the melted ice heating

$Q_1=m_w c_w (T_{w} - T)$

$Q_2=m_i c_i$

$Q_3=m_i c_w (T-T_i)$

$m_w$ - mass of the water

$m_i$ - mass of the ice

$T_w$ - initial temperature of the water

$T_i$ ​ - initial temperature of the ice

$T$ - resulting temperature of the mixture

$c_w$ - specific heat of the water, $c_w \approx 4.186 \; J/(g \cdot \degree C)$

$c_i$ - specific latent heat of fusion of the ice, $c_i \approx 334 \; J/g$

$1 \cdot 4.186 \cdot (100-T)=1 \cdot 334 + 1 \cdot 4.186 \cdot (T-0)$

$418.6-4.186T=334 + 4.186T$

$8.372T=84.6$

$T= \frac{84.6}{8.372}$

$T \approx 10 \degree C$

Answer: $10 \degree C$.