Question

$300g$ of water at $25^{\circ }C$ is added to $100g$ of ice at $0^{\circ }C$. The find temperature of mixture will be

• A. $25^{\circ }C$
• B. $0{}^{\circ }C$
• C. $12.5^{\circ }C$
• D. $30^{\circ }C$

Answer 1

Answer: (B)

Given, mass of water $m_w=300g$
Temperature of water, $T_w=25^{\circ }C$
Mass of ice, $m_i=100g$
$T_i=0^{\circ }C$
Heat required to convert the ice at $0^{\circ }C$ to water at $0^{\circ }C$ $Q_1=m_t{L}_i=100\times 80\left[L_i=80cal/g\right]$
$=8000cal\dots$. (i)
Heat given by water from $25^{\circ }C$ water to $0^{\circ }C$ water $Q_2=m_{w}c\Delta T=300\times 1\times (25-0)$
$\Rightarrow =7500cal\dots \dots .$ (ii)
From Eqs. (i) and (ii), we observed that
$\$Q_{-}\{2\}$ Hence, total ice will not be melt,
so final temperature of mixture will be $0^{\circ }C$ with some unmelted ice in mixture.