Question

A 0.400kg sample is placed in a cooling apparatus that removes energy as heat at a constant rate. The temperature T of the sample vs time t; the horiontal scale is set by ts=80.0min. The sample freezes during the energy removal. The specific heat of the sample in its initial liquid phase is 3000 J/kg *K.

a. What are the sample's heat of fusion

Answer 1

The heat of fusion is 2032.5 J/kg for the sample.

Step-by-step explanation:

The heat of fusion, also known as the latent heat of fusion, is the amount of heat energy required to change a substance from a solid to a liquid state at a constant temperature. To calculate the heat of fusion, you can use the equation:

Q = m * L

Where Q is the heat energy, m is the mass of the substance, and L is the heat of fusion. In this case, the initial liquid phase specific heat of the sample is given as 3000 J/kg * K.

Since the sample freezes during the energy removal, the heat of fusion can be calculated using the equation:

Q = m * L

where Q is the amount of heat energy, m is the mass of the sample, and L is the heat of fusion. You are given that the mass of the sample is 0.400 kg, so the heat of fusion can be calculated as:

Q = 0.400 kg * L

To solve for L, we need to rearrange the equation:

L = Q / m

Substituting the given values, the heat of fusion is:

L = 813 J / 0.400 kg = 2032.5 J/kg