Imagine a vessel containing 2.000 kg of liquid whose specific heat equals 1.50 kcal/kg/°c. Suposse that all of the liquid exists at its vaporization temperature of 450 k. We transfer 150 cal of heat energy to the liquid. What is the temperature of the liquid in the vessel after we've transferred this energy to it? help please have no idea how to solve

if it is already at its vaporization temp, the liquid cannot go higher regardless how much heat is added, it changes to vapor, and the liquid stays at the bp.

To solve this problem, we can use the equation Q = mcΔT, which relates the amount of heat energy (Q), mass (m), specific heat (c), and the change in temperature (ΔT).

Here's how we can use this equation to find the temperature of the liquid in the vessel:

1. Convert the given values to the appropriate units:
- The specific heat is given in kcal/kg/°C, so we need to convert it to cal/g/°C by dividing it by 1000.
- The heat energy transferred is given in calories, so we can use it directly.
- The mass is given in kilograms, so we need to convert it to grams by multiplying it by 1000.

2. Substitute the values into the equation:
Q = mcΔT
150 cal = (2.000 kg * 1000 g/kg) * (1.50 kcal/kg/°C / 1000 cal/g/°C) * ΔT

3. Simplify the equation:
150 cal = (2,000,000 g) * (1.50 cal/g/°C) * ΔT

4. Solve for ΔT:
ΔT = 150 cal / (2,000,000 g * 1.50 cal/g/°C)

5. Perform the calculation:
ΔT = 0.05 °C

6. Find the temperature of the liquid after transferring the heat energy:
The initial temperature of the liquid is 450 K, and the change in temperature is 0.05 °C. Therefore, the final temperature is:
Final temperature = 450 K - 0.05 °C

Please note that we can use the Kelvin scale directly because the change in temperature is small, and the difference between Celsius and Kelvin scales is negligible in this case.

By following these steps, the final temperature of the liquid in the vessel after transferring the heat energy will be approximately 449.95 K.