Question

$5gm$ of steam at ${100}^{o}C$ is passed into calorimeter containing liquid , Temperature of liquid rises from ${32}^{o}C$ to ${40}^{o}C$. Then water equivalent of calorimeter and content is

• A. $40$ gram
• B. $375$ gram
• C. $300$ gram
• D. $160$ gram

Answer 1

The correct option is B $375$ gram

Latent heat of vaporization of water = $2260kJ/Kg$

The specific heat capacity of water = $4185.5J/Kg$

Heat lost by steam = heat gained by the water and calorimeter

Formula :

Heat gained by water = mcФ

m = mass of water

c = specific heat capacity of water.

$Ф=Changeintemperature.=40-32=8$

Heat lost by steam = mLv + mcФ

Lv = latent heat of vaporisation

m = mass of steam.

$Ф=100-40=60$

Doing the substitution :

$2260000\times 0.005+60\times 0.005\times 4185.5=m\times 4185.5\times 8$

$12555.65=33484m$

$m=\frac{12555.65}{33484}=0.37497kg$

$=0.37497\times 1000=374.97kg$

$=374.97kg$