Question

Imagine you start with a glass that has 36g of solid ice. The initial temperature of the ice is 0 degree Celsius. If you leave this glass in room that has a constant temperature of 25 degree Celsius, the ice will spontaneously melt and you will end up with liquid water at 25degree Celsius...
I) Given that deltafusionH = 6.02KJ/mol and the heat capacity for water is 4.184J/g degree Celsius,what is the change in how much heat flow for system for this process(in KJ)

Answer 1

To calculate the total heat flow for the process of melting 36 g of ice at 0 °C and bringing the resulting water to 25 °C, we need to consider two steps:

1. The heat required to melt the ice at 0 °C.
2. The heat required to raise the temperature of the resulting water from 0 °C to 25 °C.

### I. Heat required to melt the ice

The heat required to melt the ice is given by the formula:

\[
q_{\text{fusion}} = n \cdot \Delta H_{\text{fusion}}
\]

where:
- \( n \) is the number of moles of ice, and
- \( \Delta H_{\text{fusion}} \) is the heat of fusion.

First, we need to calculate the number of moles of ice. The molar mass of water (H₂O) is approximately 18.015 g/mol. The number of moles of 36 g of ice is calculated as follows:

\[
n = \frac{m}{M} = \frac{36 \, \text{g}}{18.015 \, \text{g/mol}} \approx 2.00 \, \text{mol}
\]

Now we can calculate the heat required to melt the ice:

\[
q_{\text{fusion}} = n \cdot \Delta H_{\text{fusion}} = 2.00 \, \text{mol} \times 6.02 \, \text{kJ/mol} = 12.04 \, \text{kJ}
\]

### II. Heat required to raise the temperature of water

The heat required to change the temperature of the water is calculated using the formula:

\[
q_{\text{heating}} = m \cdot C \cdot \Delta T
\]

where:
- \( m \) is the mass of the water (which is still 36 g after melting),
- \( C \) is the specific heat capacity of water (4.184 J/g°C),
- \( \Delta T \) is the change in temperature (25 °C - 0 °C = 25 °C).

Substituting the values:

\[
q_{\text{heating}} = 36 \, \text{g} \times 4.184 \, \text{J/g°C} \times 25 \, \text{°C} = 36 \times 4.184 \times 25 = 3786 \, \text{J}
\]

Now convert this value from Joules to kilojoules:

\[
q_{\text{heating}} = \frac{3786 \, \text{J}}{1000} = 3.786 \, \text{kJ}
\]

### Total Heat Flow

Now we add the heat absorbed for both processes:

\[
q_{\text{total}} = q_{\text{fusion}} + q_{\text{heating}} = 12.04 \, \text{kJ} + 3.786 \, \text{kJ} = 15.826 \, \text{kJ}
\]

Thus, the total heat flow for the system during this process is **approximately 15.83 kJ**.