Question

A $10\text{kg}$ iron bar (specific heat $0.11{\text{cal/g}}^{\circ }\text{C}$) at ${80}^{\circ }\text{C}$ is placed on a big block of ice at $0^{\circ }\text{C}$. How much ice will it melts?
(Take latent heat of fusion as $80\text{cal/g}$)

• A. $1\text{kg}$
• B. $1.1\text{kg}$
• C. $2.2\text{kg}$
• D. $2\text{kg}$

Answer 1

The correct option is B $1.1\text{kg}$

Iron bar will release heat till its temperature falls to $0^{\circ }\text{C}$. After that thermal equilibrium will be reached and ice will not melt further. If $m$ be the mass of ice melted, then
Heat loss by iron bar = Heat gained by ice
$m_i{c}_i\Delta T=m\times L_{\text{ice}}$
Where,
Mass of iron bar, $m_i=10\times {10}^3\text{g}$
Specific heat of iron, $c_i=0.11{\text{cal/g}}^{\circ }\text{C}$
Latent heat of fusion, $L_{\text{ice}}=80\text{cal/g}$
$\therefore 10\times {10}^3\times 0.11\times 80=m\times 80$
$\therefore m=1.1\text{kg}$