Question

The initial temperature of 150 g of ice is ????20°C. The spe- cific heat capacity of ice is 0.5 cal/g·C° and water’s is 1 cal/g·C°. The latent heat of fusion of water is 80 cal/g. a. How much heat is required to raise the ice to 0°C and completely melt the ice? b. How much additional heat is required to heat the water (obtained by melting the ice) to 25°C? c. What is the total heat that must be added to convert the 80 g of ice at ????20°C to water at ????25°C? d. Can we find this total heat simply by computing how much heat is required to melt the ice and adding the amount of heat required to raise the temperature of 80 g of water by 45°C? Explain.

Answer 1

1. 13,500 cal

First of all, we need to find the amount of heat needed to raise the temperature of the ice from -20°C to 0°C. This is given by

`Q_1 = m C_i \Delta T`

where

m = 150 g is the mass of the ice

C_i = 0.5 cal/g·C° is the specific heat capacity of the ice

`\Delta T=0 C-(-20 C)=20^(\circ)C` is the change in temperature of the ice

Substituting,

`Q_1 = (150 g)(0.5 cal/gC)(20 C)=1500 cal`

Now we have to find the amount of heat needed to melt the ice, which is

`Q_2 = m \lambda_f`

where

m = 150 g is the mass of the ice

`\lambda_f = 80 cal/g` is the latent heat of fusion

Substituting,

`Q_2 = (150 g)(80 cal/g)=12,000 cal`

So the total heat required is

`Q_3 = 1500 cal + 12,000 cal = 13,500 cal`

2. 3750 cal

The additional amount of heat required to heat the water to 25°C is

`Q_4 = m C_w \Delta T`

where

m = 150 g is the mass of water

C_w = 1 cal/g·C is the speficic heat capacity of water

`\Delta T=25 C-0 C=25^(\circ)C` is the change in temperature

Substituting,

`Q_4 = (150 g)(1 cal/gC)(25 C)=3,750 cal`

3. 9200 cal

First of all, we need to find the amount of heat needed to raise the temperature of the ice from -20°C to 0°C. As at point 1., this is given by

`Q_1 = m C_i \Delta T`

where

m = 80 g is the mass of the ice

C_i = 0.5 cal/g·C° is the specific heat capacity of the ice

`\Delta T=0 C-(-20 C)=20^(\circ)C` is the change in temperature of the ice

Substituting,

`Q_1 = (80 g)(0.5 cal/gC)(20 C)=800 cal`

Now we have to find the amount of heat needed to melt the ice:

`Q_2 = m \lambda_f`

where

m = 80 g is the mass of the ice

`\lambda_f = 80 cal/g` is the latent heat of fusion

Substituting,

`Q_2 = (80 g)(80 cal/g)=6,400 cal`

Finally, the amount of heat required to heat the water to 25°C is

`Q_3 = m C_w \Delta T`

where

m = 80 g is the mass of water

C_w = 1 cal/g·C is the speficic heat capacity of water

`\Delta T=25 C-0 C=25^(\circ)C` is the change in temperature

Substituting,

`Q_3 = (80 g)(1 cal/gC)(25 C)=2,000 cal`

So the total heat required is

`Q=Q_1+Q_2+Q_3=800 cal+6,400 cal+2,000 cal=9,200 cal`

4. No

Step-by-step explanation:

The total heat required for this process consists of 3 different amounts of heat:

1- The heat required to bring the ice at melting temperature

2- The heat required to melt the ice, while its temperature stays constant

3- The heat required to raise the temperature of the water

However, computing how much heat is required to melt the ice and adding the amount of heat required to raise the temperature of 80 g of water by 45°C is not equivalent: in fact, the calculation of point 1) requires to use the specific heat capacity of ice, not that of water, therefore the two are not equivalent.