Question

50 g ice at $0^{o}C$ in insulator vessel, 50 g water of ${100}^{o}C$ is mixed in it, then final temperature the mixture is (neglect the heat loss)

• A. ${10}^{o}C$
• B. $0^o<<T_m<{20}^{o}C$
• C. ${20}^{o}C$
• D. Above ${20}^{o}C$

Answer 1

Answer: (A)

${10}^{o}C$

Given: mass of ice, $m_i=50g$

mass of water, $m_w=50g$

temperature of ice, $T_i=0^{o}C$

temperature of water, $T_w={100}^{o}C$

We have,

specific heat of water, $c_w=1cal/g{/}^{o}C$

latent heat of ice, $L_w=89cal/g$

Let $T^{o}C$ be the final temperature of mixture.

Now energy balance equation:

Heat gained by ice to change itself into water $+$ heat gained by melted ice(water) to raise its temperature at $T^{o}C=$ heat lost by water to reach at $T^{o}C$

$m_i{L}_i+m_i{c}_w(T-0)=m_w{c}_w(100-T)$

$50\times 80+50\times 1(T-0)=50\times 1(100-T)$

$4000+50T=5000-50T$

$100T=5000-4000=1000$

$T=\frac{1000}{100}={10}^{o}C$