Question

In a calorimeter of water equivalent 20 g, water of mass 1.1 kg is taken at 288 K temperature 373 K is passed through it and temperature of water increases by ${6.5}^{\circ }C$ then the mass of steam condensed is

• A. 17.5 g
• B. 11.7 g
• C. 15.7 g
• D. 18.2 g

Answer 1

The correct option is B 11.7 g

Let x be the mass of steam condensed. The heat obtained calorimeter with water is equal the heat from condensing x kg of steam and the heat from cooling x kg of water to the final temperature of calorimeter with water

$\begin{array}{c}x.L+x.c.\left(T_3-T_2\right)=\left(m+m_c\right).c.\left(T_2-T_1\right),\\ x=\frac{\left(m+m_c\right).c.\left(T_2-T_1\right)}{L.c.\left(T_3-T_2\right)}\\ x=\frac{\left({\mathrm{1.1.10}}^{3}g+20g\right).1\frac{cal}{g.degree}.6.5ree}{540\frac{cal}{g}+1\frac{cal}{g.ree}.\left(373k-294.5k\right)}\\ =12g=0.12kg\end{array}$