Question

Find the quantity of heat required to convert 40 g of ice at `-20^(@)`C into water at `20^(@)`C. Given `L_(ice) = 0.336 xx 10^(6) J/kg.` Specific heat of ice = 2100 J/kg-K
Specific heat of water = 4200 J/kg-K

Answer 1

Given, mass m=40g=`40 xx 10^(-3)`kg = 0.04kg
Temperature `T_(1)= -20^(@)C` and `T_(2)=20^(@)C`
Latent heat of ice `L_(ice) = 0.366 xx 10^(6)` J/kg
Specific heat of water `c_(w)` = u4200 J/kg-K
Heat required to conveert the ice into water at `0^(@)`C.
`mL_(ice)` = `0.04 xx 0.336 xx 10^(6)`=13440 J
Heat required to heat water from `0^(@)`C To `10^(@)`C
`=mc_(w)T_(2)` = `0.04 xx 4200 xx 20`=3360 J
Total heat required = 1680 + 13440 +3360 = 18480 J
Note: While doing numericals, keep in mind whenever there is a change in temperature, use Q = mc`DeltaT` or Q = `nCDeltaT`. If there is a change in state, use Q = mL