Question

$80\mathrm{g}$ of water at ${10}^{°}\mathrm{C}$ is mixed with steam at${100}^{°}\mathrm{C}$. The amount of water present after it acquires a temperature of ${40}^{°}\mathrm{C}$ is?

Answer 1

Step 1: Given data

Mass of water at ${10}^{°}\mathrm{C}$$=80\mathrm{g}$

The temperature of steam at which water is mixed$={100}^{°}\mathrm{C}$

The temperature at which amount of water is to be calculated$={40}^{°}\mathrm{C}$

Step 2: Calculating the amount of water at ${40}^{°}\mathrm{C}$

specific heat of water ${\mathrm{s}}_{\mathrm{w}}=1\mathrm{cal}{\mathrm{g}}^{-1}{}^{°}\mathrm{C}^{-1}$

Latent heat of steam ${\mathrm{L}}_{\mathrm{s}}$$=540\mathrm{cal}{\mathrm{g}}^{-1}$

Heat lost by “m” gram of steam at ${100}^{°}\mathrm{C}$ to change into the water at ${40}^{°}\mathrm{C}$ is

${\mathrm{Q}}_1={\mathrm{mL}}_{\mathrm{s}}+{\mathrm{ms}}_{\mathrm{w}}∆{\mathrm{T}}_{\mathrm{w}}$

$=\mathrm{m}\times 540+\mathrm{m}\times 1\times (100-40)$

$=540\mathrm{m}+60\mathrm{m}\Rightarrow 600\mathrm{m}$

Heat gained by $80\mathrm{g}$ water to change from ${10}^{°}\mathrm{C}$ to ${40}^{°}\mathrm{C}$

${\mathrm{Q}}_2={\mathrm{m}}_{\mathrm{w}}{\mathrm{s}}_{\mathrm{w}}∆\mathrm{T}$

$=80\times 1\times (40-10)\Rightarrow 2400$

According to the principle of calorimetry, ${\mathrm{Q}}_1={\mathrm{Q}}_2$

$600\mathrm{m}=2400,\mathrm{m}=4\mathrm{g}$

The total mass of water present$=80\mathrm{g}+\mathrm{m}$

$=80\mathrm{g}+4\mathrm{g}\Rightarrow 84\mathrm{g}$

Therefore, the amount of water present after it acquires a temperature of ${40}^{°}\mathrm{C}$ is $84\mathrm{g}$