JEE Main & Advanced
Physics
Thermometry, Calorimetry & Thermal Expansion
Question Bank
Numerical Value Type Questions - Calorimetry
question_answer
A calorimeter contains a mixture of 250 g of water and 200 g of ice at\[0{}^\circ C\]. The water equivalent of the calorimeter is 60 g. Now 300 g of steam at \[100{}^\circ C\]is passed through this mixture, then calculate the final temperature (in\[{}^\circ C\]) of the mixture [Latent heat of steam = 536 cal/g and Latent heat of\[ice=80\text{ }cal/g\]]