Question

practice activity # 3
solve the following problem. show complete details of your answers.
how much heat is required to change 500g of water at 100 °c into vapor at 108 °c?

Explanation:

Step1: Identify the processes

We need to consider two processes: 1) Vaporization of water at \(100^\circ\text{C}\) to steam at \(100^\circ\text{C}\) (using latent heat of vaporization), and 2) Heating the steam from \(100^\circ\text{C}\) to \(108^\circ\text{C}\) (using specific heat of steam).

Step2: Recall the formulas

• Latent heat of vaporization of water, \(L = 2260\space\text{J/g}\) (or \(2.26\times 10^6\space\text{J/kg}\))
• Specific heat of steam, \(c = 2.01\space\text{J/g}^\circ\text{C}\) (or \(2010\space\text{J/kg}^\circ\text{C}\))
• Mass of water/steam, \(m = 500\space\text{g}\)

Step3: Calculate heat for vaporization (\(Q_1\))

The formula for heat during phase change (vaporization) is \(Q_1 = mL\)
Substitute \(m = 500\space\text{g}\) and \(L = 2260\space\text{J/g}\)
\(Q_1 = 500\times2260 = 1130000\space\text{J}\)

Step4: Calculate heat for heating steam (\(Q_2\))

The formula for heat during temperature change is \(Q_2 = mc\Delta T\)
Here, \(\Delta T = 108 - 100 = 8^\circ\text{C}\), \(m = 500\space\text{g}\), \(c = 2.01\space\text{J/g}^\circ\text{C}\)
\(Q_2 = 500\times2.01\times8 = 500\times16.08 = 8040\space\text{J}\)

Step5: Total heat (\(Q\))

Total heat required is \(Q = Q_1 + Q_2\)
\(Q = 1130000 + 8040 = 1138040\space\text{J}\) or \(1.13804\times 10^6\space\text{J}\) (or convert to kJ: \(1138.04\space\text{kJ}\))

Answer:

The total heat required is \(1138040\space\text{J}\) (or \(1138.04\space\text{kJ}\) or \(1.138\times 10^6\space\text{J}\))