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\[ice=80\text{ }cal/g\]]

Answer:

100.00