Question

A vessel of negligible heat capacity contains 40g of ice at 0${}^o$C. 8g of steam at 100${}^o$C is passed into the ice to melt it. Find the final temperature of mixture (Latent heat of ice $=336J{g}^{-1}$, Latent heat of steam $=2268J{g}^{-1}$. Sp. heat capacity of water $=4.2J{g}^{-10}{C}^{-1}$)

Answer 1

Let the final temperature of mixture be T${}^o$C, then Heat given by steam = Heat taken by ice
$m_1{L}_1+m{S}_1{t}_1=m_2{L}_2+m_2{S}_2{t}_2$, where $t_1$ and $t_2$ indicate change in temperatures.
$\Rightarrow 8\times 2268+8\times 4.2\times (100-T)=40\times 336+40\times 4.2\times (T-0)$
$\Rightarrow 18144+3360-33.6T=13440+168T$
$\Rightarrow 201.6T=8064\Rightarrow T=\frac{8064}{201.6}=\frac{80640}{2016}={40}^{o}C$