Question

$50grams$ of ice at $0^{o}C$ is mixed with $40gm$ of water at ${60}^{o}C.$ Then the ratio of ice and water at equilibrium is

• A. $2:7$
• B. $7:2$
• C. $2:3$
• D. $2:5$

Answer 1

The correct option is B $7:2$

$\begin{array}{cc}\text{Given-}& \ast \text{we have}50g\text{ice}{0}^{\circ }C\\ & \ast \text{and}40g\text{water}{60}^{\circ }C\end{array}$

$\begin{array}{c}\text{we know latent Heat of ice to convert into water}\\ L=80cal/g\end{array}$

$\text{and specific Heat capacity of water}=1cal/g$

$\begin{array}{c}\text{Let}x\text{gram of ice get converts into water.}\\ \text{if there is some ice left in the System}\\ \text{then the temperature of the system must}\\ \text{be equal to zero.}{0}^{\circ }C\text{.}\end{array}$

$\begin{array}{c}\Rightarrow \text{Heat lost By 40 g water = Heat used in melting}x\text{gram}\\ \text{of ice}\end{array}$

$\begin{array}{c}\Rightarrow 40\times 1\times 60=x\times 80\\ \Rightarrow x=30g\end{array}$

$\begin{array}{c}\Rightarrow \text{Ice finally=}40+30g=70g\\ \text{Water lift}=50-30g=20g\\ \Rightarrow \text{Ratio of ice to water}=70:20\\ =7:2\end{array}$