Question

A copper calorimeter of mass $100gm$ contains $200gm$ of a mixture of ice and water. Steam at ${100}^{\circ }C$ under normal pressure is passed into the calorimeter and the temperature of the mixture is allowed to rise to ${50}^{\circ }C$. If the mass of the calorimeter and its contents is now $330gm$, what was the ratio of ice and water in the beginning? Neglect heat losses.
Given: Specific heat capacity of copper $=0.42\times {10}^{3}Jk{g}^{-1}{K}^{-1}$ ,
Specfic heat capacity of water $=4.2\times {10}^{3}Jk{g}^{-1}{K}^{-1}$
Specific heat of fusion of ice $=3.36\times {10}^{5}Jk{g}^{-1}$
Latent heat of condensation of steam $=22.5\times {10}^{5}Jk{g}^{-1}$

• A. $1:1.26$
• B. $1:5.26$
• C. $5:1.26$
• D. $7:1.86$

Answer 1

The correct option is A $1:1.26$

Let the mass of ice be m kg. Then mass of water is 0.2-m kg.

Heat loss by steam in converting to water and lowering temperature from ${100}^{o}C$ to ${50}^{o}C$ will be used to raise temperature of calorimeter and initial ice and water mi\timesture. First ice will convert to water and then the temperature will rise to ${50}^{o}C$

$⟹\left(0.1\right)(0.42\times {10}^3)(50-0)+m(3.36\times {10}^5)+\left(0.2\right)(4.2\times {10}^3)(50-0)=\left(0.03\right)(22.5\times {10}^5)+\left(0.03\right)(4.2\times {10}^3)(100-50)$

$⟹\left(0.18\right)(4.2\times {10}^3)\left(50\right)+m(3.36\times {10}^5)=\left(0.03\right)(22.5\times {10}^5)$

$⟹m\approx 88g$

$⟹\frac{m}{200-m}=1:1.26$