Question

A 250 g block of ice is removed from the refrigerator at -8.0°C. How much thermal energy does the ice absorb as it warms to room temperature (22°C)? The heat of fusion of water is 3.34 x 105 J/kg.

Answer 1

Answer:

Q = 114895 J

Explanation:

To find the thermal energy gained by the ice you use the following formula:

$Q=mc(T_2-T_1)+H_f\ m$

m: mass of the ice = 0.250kg

T2: final temperature = 22°C

T1: initial temperature = -8.0°C

Hf: heat of fusion of water = 3.34*10^5 J/kg

c: specific heat of water = 4186 J/kg

By replacing the values of the parameters you have:

$Q=(0.250kg)(4186J/kg\°C)(22+8)\°C+(3.34*10^5 J/kg)(0.250kg)\\\\Q=114895\ J$

where you have considered that ice melts completely