Question

A piece of ice of mass $40g$ is dropped into $200g$ of water at ${50}^{o}C$. Calculate the final temperature of the water after all the ice has melted. Specific heat capacity of water $=4200Jk{g}^{-1}{K}^{-1}$ Specific latent heat of fusion of ice $=336\times {10}^{3}Jk{g}^{-1}$

Answer 1

Mass of ice $\left(m_1\right)=40g=\frac{40}{1000}kg$
Mass of water $\left(m_2\right)=200g=\frac{200}{1000}kg$
Temperature of water ${50}^{o}C$. Let final temp. $=T^{o}C$
Heat taken by ice = Heat given by water
$m_{1}L+m_1{S}_1{T}_1=m_2{S}_2{T}_2$

$\Rightarrow \frac{40}{1000}\times 336\times {10}^3+\frac{40}{1000}\times 4.2\times {10}^3\times T$ $=\frac{200}{1000}\times 4.2\times {10}^3\times (50-T)$

$\Rightarrow 40\times 336+40\times 4.2T=200\times 4.2\times (50-T)$

$\Rightarrow 336+4.2T=21(50-T)$
$\Rightarrow 25.2T=714\Rightarrow T=\frac{714}{25.2}={28.33}^{o}C$