MCAT General Chemistry Question 208: Answer and Explanation
Home > MCAT Test > MCAT general chemistry practice tests
Test Information
- Use your browser's back button to return to your test results.
- Do more MCAT general chemistry practice tests.
Question: 208
6. A pot containing 0.5 L of water at sea level is brought to 100°C. It is insulated around its sides to minimize heat loss to the environment, and heat is applied at the bottom of the container at a rate of 6 kJ/min over 3 minutes. What is the resulting temperature of the water? (∆Hvap = 40.7 kJ/kg; c(g) = 1.9 kJ/kg • °C; c(l) = 4.2 kJ/kg • °C)
- A. 115°C
- B. 108°C
- C. 104°C
- D. 100°C
Correct Answer: D
Explanation:
D The addition of 6 kJ/min for three minutes imparts 18 kJ of heat to the sample. However, since the ∆Hvap of water is 40.7 kJ/mol, and 0.5 L is roughly 22 mol (1 L of H2O ≈ 55 mol), there is nowhere near enough heat provided to vaporize the entire sample. As such all the heat given to the sample is going toward vaporization and not toward increasing temperature. The temperature will remain constant at 100°C.