How much heat should be supplied to convert ice of mass 50 g at `0^@C` into water at `40^@ C` ?

Answer

Amount of heat energy that should be supplied is 6000 cal.

Explanation

Given : Mass of ice/water (m)=50 g
Change in temperature `(Delta T)`
`= 40-0-40 ^@C`
We know that: Latent heat of melting of ice `=L=("melt")=80 cal//g` ,Specific heat of water `C_("w")=1 cal//g^@C`
To find, Heat supplied (Q)
Formulae: i Heat required to convent ice at `0^@C` into water at `0^@C " is " Q_1=mL_("melt")`
ii. Heat required to raise temperature of water from `0^@C` to `40^@C` is
`Q_2= mxxC_w xx Delta T`
Calculation: From formula (i),
`Q_1=50 xx 1 xx (40-0)=2000` cal Total heat energy required
`Q=Q_1+Q_2`
`=4000 +2000=6000 cal`