Answer :
Final Answer:
The heat required is approximately 98.2 kJ, calculated by considering the energy needed to raise the water's temperature, convert it to steam, and raise the steam's temperature to 100.0 °C. So, the correct option is c. 98.2 kJ
Explanation:
To calculate the heat required, we first need to find the heat needed to raise the temperature of water from 67.0 °C to 100.0 °C, then the heat required to convert the water at 100.0 °C to steam at 100.0 °C, and finally the heat required to raise the temperature of steam from 100.0 °C to its final state. The total heat required is the sum of these three parts.
1. **Heating water from 67.0 °C to 100.0 °C:** Use the formula[tex]\(q = mcΔT\),[/tex] where \(m\) is the mass of water (2.30 g), \(c\) is the specific heat capacity of water (4.18 J/g°C), and ΔT is the temperature change (100.0 °C - 67.0 °C = 33.0 °C).
[tex]\[q_1 = 2.30 \text{ g} \times 4.18 \text{ J/g°C} \times 33.0 \text{ °C} = 318.42 \text{ J}\][/tex]
2. **Converting water at 100.0 °C to steam at 100.0 °C:** Use the heat of vaporization of water (2260 J/g).
[tex]\[q_2 = 2.30 \text{ g} \times 2260 \text{ J/g} = 5186 \text{ J}\][/tex]
3. **Heating steam from 100.0 °C to final state:** Use the formula \(q = mcΔT\), where \(m\) is the mass of steam (2.30 g), \(c\) is the specific heat capacity of steam (2.01 J/g°C), and ΔT is the temperature change (100.0 °C - 100.0 °C = 0 °C).
[tex]\[q_3 = 2.30 \text{ g} \times 2.01 \text{ J/g°C} \times 0 \text{ °C} = 0 \text{ J}\][/tex]
The total heat required is the sum of \(q_1\), \(q_2\), and \(q_3\):
[tex]\[q_{\text{total}} = 318.42 \text{ J} + 5186 \text{ J} + 0 \text{ J} = 5504.42 \text{ J} = 5.504 \text{ kJ} \approx 98.2 \text{ kJ}\][/tex]
Therefore, the amount of heat required is approximately 98.2 kJ.
