Question

A 3.84-kg block of ice at 0.0°C is heated until it turns into water at 68.5°C. How much heat is used in the process?

Answer 1

Answer:

Heat required to raise the temperature of ice is given as

$Q = 2.39 \times 10^6 J$

Explanation:

Heat required to raise the temperature of the object is given as

$Q = mL + ms(\Delta T)$

here we know that

m = 3.84 kg

L = 335000 J/kg[/tex]

s = 4186 J/C K

now we have

$Q = 3.84 (335000) + 3.84(4186)(68.5 - 0)$

$Q = 1.29 \times 10^6 + 1.1\times 10^6$

$Q = 2.39 \times 10^6 J$