Question

The energy (Q) required to heat an amount M of material with heat capacity Cp a temperature difference of △ T can be calculated by: The heat capacity of water is 1 cal/g/deg C. There are 4.18 J/cal and there are 454 grams per lb. Look up other conversion factors as necessary. Approximately how much energy (in kJ) is required to heat 25 lbs of water 1 degree C? Group of answer choices

Answer 1

Q = 47443 J

Step-by-step explanation:

Mas of the water is given as

`m = 25 lbs`

1 lb = 454 g

so we have

`m = 25 * 454`

`m = 11.35 kg`

specific heat capacity of the water is given as

`s = 1 cal/g C`

`s = 1000* 4.18 J/kg C`

`s = 4180 J/kg C`

now heat required is given as

`Q = 11.35 (4180) (1)`

`Q = 47443 J`