Sample Problems on Latent Heat of Fusion

Problem 1: Calculate the latent heat of fusion for a body of mass of 40 g at 20°C if it absorbs heat at 80°C. It is given that the specific latent heat of steam is 540 cal/g and the specific heat of the body is 0.5 cal/g °C.

Solution:

We have,

m = 40, ΔT = 80 – 20 = 60, L = 540 and c = 0.5

Using the formula we get,

Q = mL + mcΔt

⇒ Q = (40 × 540) + (40 × 0.5 × 60)

⇒ Q = 21600 + 1200

⇒ Q = 22800 Cal

Problem 2: Calculate the latent heat of fusion for a body of mass 20 g at 40°C if it absorbs heat at 100°C. It is given that the specific latent heat of steam is 540 cal/g and the specific heat of the body is 0.1 cal/g °C.

Solution:

We have,

m = 20, ΔT = 100 – 40 = 60, L = 540, and c = 0.1

Using the formula we get,

Q = mL + mcΔt

⇒ Q = (20 × 540) + (20 × 0.1 × 60)

⇒ Q = 10800 + 120

⇒ Q = 10920 Cal

Problem 3: Calculate the latent heat of fusion for 7 g of water getting converted into ice. It is given that the specific latent heat of ice is 80 cal/g.

Solution:

We have,

m = 7, and L = 80

Using the formula we get,

Q = mL

⇒ Q = 7 (80)

⇒ Q = 560 Cal

Problem 4: Calculate the latent heat of fusion for 60 g of steam getting converted into water. It is given that the specific latent heat of water is 533 cal/g.

Solution:

We have,

m = 60

L = 533

Using the formula we get,

Q = mL

⇒ Q = 60 (533)

⇒ Q = 31980 cal

Problem 5: Calculate the mass of water that gets converted into ice given that the specific latent heat of ice is 80 cal/g and the latent heat of fusion is 200 cal.

Solution:

We have,

Q = 200

L = 80

Using the formula we get,

Q = mL

⇒ m = Q/L

⇒ m = 200/80

⇒ m = 2.5 g

Problem 6: Calculate the mass of steam getting converted into water given that the specific latent heat of water is 533 cal/g and the latent heat of fusion is 700 cal.

Solution:

We have,

Q = 700

L = 533

Using the formula we get,

Q = mL

⇒ m = Q/L

⇒ m = 700/533

⇒ m = 1.31 g

Problem 7: Calculate the latent heat of fusion for a body of mass 30 g if its specific latent heat of steam is 540 cal/g and the heat absorbed by it is 200 calories.

Solution:

We have,

m = 30

L = 540

Q’ = 200

Using the formula we get,

Q = mL + Q’

⇒ Q = (30 × 540) + 200

⇒ Q = 16200 + 200

⇒ Q = 16400 Cal

Latent Heat of Fusion is one of the latent heats in chemistry like the latent heat of vaporization and latent heat of sublimation. When a substance is changing its phase from liquid to solid or solid to the gas of gas to a liquid, this latent heat comes in handy to find the energy of the reaction. Latent Heat of Fusion is related to the solid-to-liquid phase change. In this article, we will explore all the topics related to the latent heat of fusion and its formula. We’ll also learn about the latent heat of fusion for various different elements as well as compounds. 

Let’s start our learning about the concept with the name “Latent Heat of Fusion”.