Calculate the heat of fusion of ice from the following data of ice at `0^@C` added to water. Mass of calorimeter`=60g`, mass of calorimeter`+` water `=460g`,
mass of calorimeter` +`water`+ `ice`=618g`, initial temperature of water `=38^@C`, final temperature of the mixture`=5^@C`. The specific heat of calorimeter `=0.10 cal//g//^@C`. Assume that the calorimeter was also at `0^@`C initially

Options

A. `78.35 cal//g`
B. `58.55 cal//g`
C. `68.95 cal//g`
D. `48.35 cal//g`

Answer

A

Explanation

Mass of water`=480-60=400g`
mass of ice`=618-460=158g`
Heat lost by water`=`heat gained by ice to melt`+`heat gained by (water obtained from melting of ice`+` calorimeter) to reach `5^@C`
`implies400xx1xx(38-5)=158xxL+158xx1xx5+60xx0.1xx5`
(where L is the latent heat of fusion ice)
`impliesL=78.35 cal//g`