Question

How many calories are required to convert 17 g of ice at 0.0°C to liquid water at 32.0°C? The heat of fusion of water is 80. cal/g.

Answer 1

To convert 17 g of ice at 0.0°C to liquid water at 32.0°C, a total of 1904 calories is required. This includes the energy needed for melting the ice and then heating the resulting water.

Step-by-step explanation:

To calculate the total amount of calories needed to convert 17 g of ice at 0.0°C to liquid water at 32.0°C, we need to consider two steps: the melting of ice and the heating of the resulting water. First, we'll use the heat of fusion to calculate the energy required to melt the ice, and then we'll use the specific heat capacity of water to calculate the energy needed to raise the temperature of the resulting water to 32.0°C.

1. Heat required to melt ice (using heat of fusion):

• Q1 = mass of ice * heat of fusion
• Q1 = 17 g * 80. cal/g
• Q1 = 1360 calories

2. Heat required to raise the temperature of the melted ice (now water) from 0°C to 32.0°C:

• Q2 = mass of water * specific heat capacity of water * temperature change
• Q2 = 17 g * 1 cal/g°C * (32.0°C - 0.0°C)
• Q2 = 17 g * 1 cal/g°C * 32.0°C
• Q2 = 544 calories

The total heat required is the sum of Q1 and Q2:

• Total heat (Q) = Q1 + Q2
• Total heat (Q) = 1360 calories + 544 calories
• Total heat (Q) = 1904 calories

Answer 2

Answer : The heat required is, 1904 calories.

Explanation :

The process involved in this problem are :

`(1):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(2):H_2O(l)(0^oC)\rightarrow H_2O(l)(32.0^oC)`

The expression used will be:

`\Delta H=m* \Delta H_(fusion)+[m* c_(p,l)* (T_(final)-T_(initial))]`

where,

m = mass of ice = 17 g

`c_(p,l)` = specific heat of liquid water =
`1cal/g^oC`

`\Delta H_(fusion)` = enthalpy change for fusion =
`80.0cal/g`

Now put all the given values in the above expression, we get:

`\Delta H=17g* 80.0cal/g+[17g* 1cal/g^oC* (32.0-0)^oC]`

`\Delta H=1904cal`

Therefore, the heat required is, 1904 calories.