Question

A 15 g ice sample at -5°C is heated to water at 100°C. Calculate the ampount of heat needed. Specific heat of water and ice are 4.184 J/g °C and 2.108 J/g °C respectively. And heat of fusion is 6.01 kJ/mol

Answer 1

To warm a 15 g ice sample from -5°C to water at 100°C, the calculations for heating, melting, and then heating the water must be carried out in steps, resulting in a total heat requirement of 11,430.1 joules.

Step-by-step explanation:

To calculate the amount of heat needed to warm a 15 g ice sample from -5°C to water at 100°C, we need to take several steps, considering the specific heat of ice, the heat of fusion, and the specific heat of the water. First, we will heat the ice from -5°C to 0°C, using the specific heat of ice. Next, we melt the ice at 0°C using the heat of fusion. Finally, we heat the resulting liquid water from 0°C to 100°C, using the specific heat of water.

The calculations will be as follows:

1. Q1 (to heat the ice to 0°C):

Q1 = mass of ice × specific heat of ice × change in temperature
Q1 = 15 g × 2.108 J/g°C × (0°C - (-5°C))
Q1 = 15 g × 2.108 J/g°C × 5°C
Q1 = 158.1 J

1. Q2 (to melt the ice):

First, we convert grams to moles:
n = mass / molar mass of H2O
n = 15 g / 18.015 g/mol
n = 0.832 mol

Next, we use the heat of fusion:
Q2 = moles of ice × heat of fusion
Q2 = 0.832 mol × 6.01 kJ/mol
Q2 = 4.996 kJ = 4996 J

1. Q3 (to heat the water to 100°C):

Q3 = mass of water × specific heat of water × change in temperature
Q3 = 15 g × 4.184 J/g°C × (100°C - 0°C)
Q3 = 15 g × 4.184 J/g°C × 100°C
Q3 = 6276 J

Total heat required, Qtotal = Q1 + Q2 + Q3
Qtotal = 158.1 J + 4996 J + 6276 J
Qtotal = 11430.1 J

Thus, the total amount of heat needed is 11,430.1 joules.