Question

A calorimeter of water equivalent $20\text{g}$ contains $180\text{g}$ of water at $25{}^{\circ }\text{C}$. $m$ grams of steam at $100{}^{\circ }\text{C}$ is mixed in it till the temperature of the mixture is $31{}^{\circ }\text{C}$. The value of $m\left(\text{in g}\right)$ is close to $(\text{Latent heat of water}=540{\text{cal g}}^{-1}$, $\text{specific heat of water}=1{\text{cal g}}^{-1}{}^{\circ }{\text{C}}^{-1})$

• A. $2$
• B. $4$
• C. $3.2$
• D. $2.6$

Answer 1

The correct option is A $2$

$\text{Heat gained by water}{Q}_1=m_w{C}_w(T_{\text{mix}}-T_w)\Rightarrow Q_1=(180+20)\times 1\times (31-25)\Rightarrow Q_1=1200\text{cal}$

$\text{Heat lost by steam}{Q}_2=m{L}_{\text{steam}}+m{C}_w(T_s-T_{\text{mix}})$

$\Rightarrow Q_2=m\times 540+m\left(1\right)\times (100-31)$

$\Rightarrow Q_2=609m$

From the principal of calorimeter,
$\text{Heat lost = Heat gained}$

$\Rightarrow 1200=m\left(609\right)\Rightarrow m\approx 2$

Hence, option $\left(A\right)$ is correct.