Question

$10g$ of ice at $0^{o}C$ is mixed with $100g$ of water at ${50}^{o}C$ in a calorimeter. The final temperature of the mixture is [Specific heat of water$=1cal{g}^{-1}{}^o{C}^{-1}$, latent heat of fusion of ice$=80cal{g}^{-1}]$

• A. ${31.2}^{o}C$
• B. ${32.8}^{o}C$
• C. ${36.7}^{o}C$
• D. ${38.2}^{o}C$

Answer 1

The correct option is D ${38.2}^{o}C$

Here, Mass of water $m_w=100g$
Mass of ice, $m_i=10g$
Specific heat of water,$S_w=1cal{g}^{-1}{}^o{C}^{-1}$
Latent heat of fusion of ice,$L_{fi}=80cal{g}^{-1}$

Let $T$ be the final temperature of the mixture.

Amount of heat lost by water
$=m_w{s}_w(△T{)}_w=100\times 1\times (50-T)$

Amount of heat gained by ice
$=m_i{L}_{fi}+m_i{s}_w(△T{)}_i=10\times 80+10\times 1\times (T-0)$

According to principle of calorimetry:
Heat lost = Heat gained
$100\times 1\times (50-T)=10\times 80+10\times 1\times (T-0)$
$500-10T=80+T$
$11T=420orT={38.2}^{o}C$