Entropy of melting
A block of a certain substance (m = 0.3 kg) sits in a cold room at its melting point 170 K. A lab demonstrator supplies heat at a rate of 200 W for 140 seconds by focusing light on the block to (just) completely melt the block.
(a) What is the latent heat of fusion of the substance?
(b)What is the change in entropy of the block due to the melting process?
Cheers
(a) Latent H.O.F. = Q
= (energy added during melting)/(mass)
(b) Entropy Change = Q/T
= (Energy added)/(170 K)
To calculate the latent heat of fusion of the substance, we can use the formula:
Q = mL
where Q is the amount of heat supplied, m is the mass of the substance, and L is the latent heat of fusion.
In this case, we are given the heat supplied (200 W) and the time the heat is supplied (140 seconds), so we can calculate Q as:
Q = P * t = 200 W * 140 s
Next, we need to convert the given mass of the substance (0.3 kg) into grams:
m = 0.3 kg * 1000 g/kg
Now that we have the values for Q and m, we can rearrange the formula to solve for L:
L = Q / m
Substituting the values, we get:
L = (200 W * 140 s) / (0.3 kg * 1000 g/kg)
Simplifying the expression, we find:
L = 93333.33 J/g
Therefore, the latent heat of fusion of the substance is 93333.33 J/g.
To calculate the change in entropy of the block due to the melting process, we can use the formula:
ΔS = Q / T
where ΔS is the change in entropy, Q is the amount of heat supplied, and T is the temperature at which the heat is supplied.
In this case, the temperature at which the heat is supplied is the melting point of the substance, which is 170 K.
We already have the value for Q from the previous calculation, so now we can calculate ΔS as:
ΔS = (200 W * 140 s) / 170 K
Simplifying the expression, we find:
ΔS = 1647.06 J/K
Therefore, the change in entropy of the block due to the melting process is 1647.06 J/K.