Question

A block of ice weighing 20 lb is taken from the freezer where it was stored at -15"F. How many Btu of heat will be required to convert the ice to water at 200°F?

Answer 1

Heat required =7126.58 Btu.

Step-by-step explanation:

Given that

Mass m=20 lb

We know that

1 lb =0.45 kg

So 20 lb=9 kg

m=9 kg

Ice at -15° F and we have to covert it at 200° F.

First ice will take sensible heat at up to 32 F then it will take latent heat at constant temperature and temperature will remain 32 F.After that it will convert in water and water will take sensible heat and reach at 200 F.

We know that

Specific heat for ice
`C_p=2.03\ KJ/kg.K`

Latent heat for ice H=336 KJ/kg

Specific heat for ice
`C_p=4.187\ KJ/kg.K`

We know that sensible heat given as

`Q=mC_p\Delta T`

Heat for -15F to 32 F:

`Q=mC_p\Delta T`

`Q=9* 2.03(32+15) KJ`

Q=858.69 KJ

Heat for 32 Fto 200 F:

`Q=mC_p\Delta T`

`Q=9* 4.187(200-32) KJ`

Q=6330.74 KJ

Total heat=858.69 + 336 +6330.74 KJ

Total heat=7525.43 KJ

We know that 1 KJ=0.947 Btu

So 7525.43 KJ=7126.58 Btu

So heat required to covert ice into water is 7126.58 Btu.