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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

Respuesta :

Answer:

Q = 47443 J

Explanation:

Mas of the water is given as

[tex]m = 25 lbs[/tex]

1 lb = 454 g

so we have

[tex]m = 25 \times 454[/tex]

[tex]m = 11.35 kg [/tex]

specific heat capacity of the water is given as

[tex]s = 1 cal/g C[/tex]

[tex]s = 1000\times 4.18 J/kg C[/tex]

[tex]s = 4180 J/kg C[/tex]

now heat required is given as

[tex]Q = 11.35 (4180) (1)[/tex]

[tex]Q = 47443 J[/tex]

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