Question

A student (with slow reflexes) puts his hand in a stream of steam at \(100 .^{\circ} \mathrm{C}\) until \(1.5 \mathrm{~g}\) of water have condensed. If the water then cools to room temperature \(\left(20.0^{\circ} \mathrm{C}\right)\), how many joules have been absorbed by the student's hand?

Short answer

3891.6 J

Step-by-step solution

- Identify the Processes

There are two processes: condensation of steam to liquid water, and cooling the water down to room temperature.

- Calculate Heat of Condensation

Use the heat of vaporization of water which is given as the latent heat of condensation, \( L_v = 2260 \, \text{J/g} \). For \( 1.5 \, \text{g} \) of steam condensing, the heat released: \( Q_1 = m \times L_v = 1.5 \, \text{g} \times 2260 \, \text{J/g} = 3390 \, \text{J} \).

- Calculate Heat Lost During Cooling

The heat lost as water cools down from \( 100^{\circ} \text{C} \) to \( 20.0^{\circ} \text{C} \) can be found using the specific heat capacity of water \( c = 4.18 \, \text{J/g} \, ^{\circ}\text{C} \) and the temperature change \( \Delta T = 100^{\circ}\text{C} - 20^{\circ}\text{C} = 80^{\circ}\text{C} \). Thus, \( Q_2 = m \times c \times \Delta T = 1.5 \, \text{g} \times 4.18 \, \text{J/g} \, ^{\circ}\text{C} \times 80^{\circ}\text{C} = 501.6 \, \text{J} \).

- Total Heat Absorbed

The total heat absorbed by the student's hand is the sum of the heat released during condensation and the heat lost during cooling. Therefore, \( Q_{\text{total}} = Q_1 + Q_2 = 3390 \, \text{J} + 501.6 \, \text{J} = 3891.6 \, \text{J} \).

Key concepts

Latent Heat of Condensation

Latent heat of condensation refers to the heat released when a gas changes back into a liquid state. In this problem, steam at 100°C condenses into water at the same temperature. The latent heat of vaporization (which is the same as the latent heat of condensation but in the opposite direction) for water is given as 2260 J/g. This value represents the energy released per gram of steam condensed.
When 1.5 g of steam condenses, we use the formula:
\[ Q_1 = m \times L_v \]
where \(m = 1.5 \, \text{g}\) and \(L_v = 2260 \, \text{J/g}\). Plugging in the numbers:
\[ Q_1 = 1.5 \, \text{g} \times 2260 \, \text{J/g} = 3390 \, \text{J} \]
This means 3390 joules of energy are released during the condensation process and subsequently absorbed by the student’s hand.

Specific Heat Capacity

The specific heat capacity is the amount of heat required to change the temperature of one gram of a substance by one degree Celsius. In this context, we use water's specific heat capacity, which is 4.18 J/g°C.
After the steam condenses into water at 100°C, the water cools down to room temperature (20°C). The formula to calculate the heat lost due to this cooling is:
\[ Q_2 = m \times c \times \Delta T \]
where \(m = 1.5 \, \text{g}\), \(c = 4.18 \, \text{J/g}\, ^{\circ}\text{C}\), and \( \Delta T = T_{\text{final}} - T_{\text{initial}}\).
Here, \( T_{\text{final}} = 20^{\circ}\text{C}\) and \( T_{\text{initial}} = 100^{\circ}\text{C}\), so:
\[ \Delta T = 100^{\circ}\text{C} - 20^{\circ}\text{C} = 80^{\circ}\text{C} \]
Substituting these values in:
\[ Q_2 = 1.5 \, \text{g} \times 4.18 \, \text{J/g}\, ^{\circ}\text{C} \times 80^{\circ}\text{C} = 501.6 \, \text{J} \]
This is the heat lost as the water cools down to room temperature.

Temperature Change

The student’s hand absorbs heat from both the condensation of steam and the subsequent cooling of water. Temperature change is an essential component of such calculations since the amount of heat energy is directly proportional to the temperature difference. We observe two temperature-related processes here:
1. Steam at 100°C condenses, but its temperature doesn't change during this phase change. However, the state changes from gas to liquid, releasing latent heat.
2. The newly formed water cools from 100°C to 20°C, a temperature change of 80°C. This affects the heat absorbed by the student’s hand as calculated by specific heat capacity.
Combining these two processes, the total heat absorbed can be calculated by adding the heat from condensation and the heat from cooling:
\[ Q_{\text{total}} = Q_1 + Q_2 = 3390 \, \text{J} + 501.6 \, \text{J} = 3891.6 \, \text{J} \]
Therefore, the total energy absorbed by the student's hand is 3891.6 joules.