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In a calorimeter of water equivalent 20 g ,water of mass 1.1 kg at 288 k temperature .if steam at teperature 373 k is passed through it and temperature of water increase by 6.5 degree c then the massss of steam condensed is

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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

 

x.L+x.c.(T_3-T_2)=(m+m_c).c.(T_2-T_1)

 

x= \frac{(m+m_c).c.(T_2-T_1)}{L+c.(T_3-T_2)}\\\\ \\ x=\frac{(1.1*10^3g+20g.1\frac{cal}{g.degree}.6.5 degree)}{540\frac{cal}{g}+1\frac{cal}{g.degree}.(373k-294.5K)}= 12g=0.12kg

 

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Safeer PP

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