Question

How much heat must be removed from 456 g of water at 25.0°C to change it into ice at - 10.0°C ? The specific heat of ice is 2090 J/(kg.K) and the latent heat of fusion of water is 33.5 x 10^4 J/kg .

Answer 1

Answer:209.98 kJ

Step-by-step explanation:

mass of water
`m=456 gm`

Initial Temperature of Water
`T_i=25^(\circ)C`

Final Temperature of water
`T_f=-10^(\circ)C`

specific heat of ice
`c=2090 J/kg-K`

Latent heat
`L=33.5* 10^4 J/kg`

specific heat of water
`c_(water)=4.184 KJ/kg-K`

Heat require to convert water at
`T=25^(\circ)C` to
`T=0^(\circ)C`

`Q_1=0.456* 4.184* (25-0)=47.69 kJ`

Heat require to convert water at
`T=0^(\circ)` to ice at
`T=0^(\circ)`

`Q_2=m* L=0.456* 33.5* 10^4=152.76 kJ`

heat require to convert ice at
`T=0^(\circ) C\ to\ T=-10^(\circ) C`

`Q_3=0.456* 2090* (0-(-10))=9.53 kJ`

Total heat
`Q=Q_1+Q_2+Q_3`

`Q=47.69+152.76+9.53=209.98 kJ`