Ask Questions, Get Answers

Want to ask us a question? Click here
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Thermodynamics
0 votes

A calorimeter (water equivalent = 25 g ) contains of 265 g of water at $\;30^{0}C\;$ . Steam is passed at $\;100^{0}C\;$ is passed which increases the temperature to $\;40^{0}C\;$ & the mass of the calorimeter and its contents by 2g . Given that the specific heat capacity of water is $\;1\;{calg^{-1}}^{0}C\;$ , the latent heat of vapourization of water is -


Can you answer this question?

1 Answer

0 votes
Answer : (a) $\;1390\;calg^{-1}$
Explanation :
Mass of steam condensed = 2g
Heat lost in condensation = $\;Q_{L_{1}}=(2)L\;cal\;$
Heat lost when water cools from $\;100^{0}C\;$ to $\;40^{0}C\;$ =$\;2 \times 1 \times (100-40)=120 cal =Q_{L_{2}}$
Heat gained by water at $\;30^{0}C\;$ to rise upto $\;40^{0}C=Q_{g}=(265+25) \times (40-30)=2650\;cal\;$
Since $\;Q_{g}=Q_{L_{1}+Q_{L_{2}}} $
$L=1390 \;calg^{-1}\;.$
answered Mar 12, 2014 by yamini.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App