- #1

takando12

- 123

- 5

## Homework Statement

5 g of water at 30°C and 5 g of ice at -20°C are mixed together in a calorimeter Find the final temperature of the mixture. Water equivalent of the calorimeter is negligible,specific heat of ice=0.5 cal/g°C and latent heat of ice =80 cal/g.

## Homework Equations

C= Q/mΔT

L=Q/m where L-----> Latent heat

## The Attempt at a Solution

The heat lost by the water is gained by the ice.But I don't really know what to do with the formula without knowing the temp change.

The Heat req. to melt the ice is Q=Lm= 400 cal. So should I equate this to CmΔT of water?

Please give me a hint as to how to proceed with this problem. And also I don't understand what the "Water equivalent of the calorimeter is negligible" is supposed to imply.

Last edited: