5400kJ of heat energy was lost when some amount of steam condensed to water for drinking purposes at 15º C…

1. Analyze the Question

We are given the total heat energy lost ($Q_{total} = 5400 \text{ kJ} = 5,400,000 \text{ J}$) during a process where steam at $100^\circ\text{C}$ condenses into water and then cools down to $15^\circ\text{C}$. We need to find the mass ($m$) of the water collected.

2. Explain the Concept

The total heat lost involves two stages:

1. Phase Change (Latent Heat): The steam at $100^\circ\text{C}$ turns into water at $100^\circ\text{C}$. The formula is $Q_1 = mL_v$. (Note: The question provides $L_f$, which usually denotes latent heat of fusion, but in the context of steam condensing, we use the latent heat of vaporization, $L_v = 2.26 \times 10^6 \text{ J/kg}$.)
2. Temperature Change (Sensible Heat): The water cools from $100^\circ\text{C}$ to $15^\circ\text{C}$. The formula is $Q_2 = mc\Delta\theta$.

Total heat $Q_{total} = Q_1 + Q_2 = m(L_v + c\Delta\theta)$.

3. Show the Solution

• Step 1: Identify the variables • $Q_{total} = 5,400,000 \text{ J}$ • $L_v = 2.26 \times 10^6 \text{ J/kg}$ • $c_w = 4200 \text{ J/kgK}$ • $\Delta\theta = 100^\circ\text{C} - 15^\circ\text{C} = 85^\circ\text{C}$

• Step 2: Set up the equation $5,400,000 = m(2,260,000 + (4200 \times 85))$

• Step 3: Calculate the values inside the bracket $4200 \times 85 = 357,000$ $2,260,000 + 357,000 = 2,617,000$

• Step 4: Solve for $m$ $m = \frac{5,400,000}{2,617,000}$ $m \approx 2.0634 \text{ kg}$

4. Identify the Answer

Rounding to two decimal places, we get $2.06 \text{ kg}$. This matches option D.

5. Tip

When dealing with phase changes, always check if the substance is changing temperature after the phase change. If it is, you must add the sensible heat ($mc\Delta\theta$) to the latent heat ($mL$). Don't forget to convert kilojoules (kJ) to joules (J) by multiplying by 1000 before starting your calculation!