Question

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

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

Answer 1

The correct option is A

$2$

Answer: (A)

Explanation for correct option

(a) Heat lost by $\text{'}\mathrm{m}\text{'}$grams of steam is gained by calorimeter and water in it.

$$\begin{gathered} (\mathrm{Steam}\mathrm{at}100°\mathrm{C})\stackrel{\mathrm{Condensation}}{\to }\left(\mathrm{Water}\mathrm{at}100°\mathrm{C}\right)\stackrel{\mathrm{Cooling}}{\to }\left(\mathrm{Water}\mathrm{at}31°\mathrm{C}°\right) \\ (\mathrm{Calorimeter}\mathrm{and}\mathrm{water}\mathrm{at}25°\mathrm{C})\stackrel{\mathrm{Heating}}{\to } \\ (\mathrm{Calorimeter}\mathrm{and}\mathrm{water}\mathrm{at}31°\mathrm{C}) \\ \mathrm{Heat}\mathrm{lost}\mathrm{by}\mathrm{steam}, \\ ={\mathrm{mL}}_1+\mathrm{m}\times {\mathrm{S}}_{\mathrm{w}}\times \left(100-31\right) \\ =\mathrm{m}[{\mathrm{L}}_1+{\mathrm{S}}_{\mathrm{w}}\left(100-31\right)] \\ =\mathrm{m}[540+1\times \left(100-31\right)]=\mathrm{m}\times 609 \\ \mathrm{Heat}\mathrm{gained}\mathrm{by}\mathrm{calorimeter}\mathrm{and}\mathrm{water}\mathrm{in}\mathrm{calorimeter}, \\ =\left(\mathrm{M}+\mathrm{W}\right)\times \mathrm{S}\times \left(31-25\right) \\ =(180+20)\times 1\times 6=200\times 6 \\ \mathrm{As},\mathrm{heat}\mathrm{lost}=\mathrm{heat}\mathrm{gained} \\ \mathrm{m}\times 609=200\times 6 \\ \mathrm{m}=\frac{200\times 6}{609}=\frac{1200}{609}=1.97\approx \mathbf{2}\mathbf{g} \end{gathered}$$

Hence option (A) is the correct answer.