Question

"If the specific latent heat of fusion of ice is 336 J g- and the specific heat capacity of water is 4.2 Jg -1 °C-1, how much heat is needed to change 20 g of ice at 0 °C to water at 25 °C?"

A. 25,200 J
B. 16,800 J
C. 18,480 J
D. 13,440 J

Answer 1

The total heat needed to change 20 g of ice at 0 °C to water at 25 °C is 6720 J for the melting process plus 52500 J for heating the resulting water, summing up to 59220 J. This correct answer is not listed among the provided options A, B, C, or D.

Step-by-step explanation:

To calculate the heat needed to change 20 g of ice at 0 °C to water at 25 °C, we need to consider two processes: melting the ice and then warming the water. The specific latent heat of fusion of ice is given as 336 J g-1, which means 336 J of heat is required to melt every gram of ice.

For the melting process:

• Heat needed to melt ice (Qmelting) = mass of ice (m) × latent heat of fusion (Lf)
• Qmelting = 20 g × 336 J g-1 = 6720 J

After melting, the water at 0 °C needs to be heated to 25 °C, for which we use the specific heat capacity of water, 4.2 J g-1 °C-1.

For heating the water:

• Heat needed to raise temperature (Qheating) = mass of water (m) × specific heat capacity (c) × temperature change (ΔT)
• Qheating = 20 g × 4.2 J g-1 °C-1 × 25 °C = 2100 J × 25 = 52500 J

Add the two amounts of heat together to find the total heat required:

• Total heat (Qtotal) = Qmelting + Qheating
• Qtotal = 6720 J + 52500 J = 59220 J

The correct option that corresponds to the calculated heat is not provided in the given choices A, B, C, or D.