. How much thermal energy must be removed from a 0.25-kg chunk of ice to lower its temperature by 18C ? The specific heat of ice to is 2100 J/kg ⋅K and the heat of fusion for ice is 334×10^3J/kg.

Answer 1

146475

Answer 2

The calculation involves two steps:

1. First, we need to calculate the amount of heat required to lower the temperature of the ice.

Q1 = m * c * ΔT

where Q1 is the amount of heat, m is the mass of the ice (0.25 kg), c is the specific heat of ice (2100 J/kg ⋅K), and ΔT is the change in temperature (-18C).

Q1 = 0.25 kg * 2100 J/kg ⋅K * (-18C) = -94500 J

Note that the value of ΔT is negative because the temperature is being lowered.

2. Second, we need to calculate the amount of heat required to melt the ice.

Q2 = m * Lf

where Q2 is the amount of heat, m is the mass of the ice (0.25 kg), and Lf is the heat of fusion for ice (334×10^3 J/kg).

Q2 = 0.25 kg * 334×10^3 J/kg = 83,500 J

Note that the value of Lf is positive because the ice is melting.

Therefore, the total amount of thermal energy that must be removed from the ice is:

Q = Q1 + Q2 = -94500 J + 83500 J = -11,000 J

The negative sign indicates that the thermal energy is being removed from the ice. The answer is approximately -11,000 J.