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How much heat is required to convert 422 g of liquid H₂O at 23.5 °C into steam at 150 °C?

a) 4,930 J
b) 6,540 J
c) 8,210 J
d) 9,420 J

User BinaryDi
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1 Answer

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Final answer:

To find the heat required to turn 422 g of water at 23.5 °C into 150 °C steam, calculate the heat to increase temperature to 100 °C, then add the heat needed for vaporization.

Step-by-step explanation:

To calculate the amount of heat required to convert 422 g of liquid H₂O at 23.5 °C into steam at 150 °C, we need to take into account two processes: heating the water from 23.5 °C to 100 °C and vaporizing it at 100 °C. We'll use the specific heat of water and the heat of vaporization.


First, determine the heat required to raise the temperature from 23.5 °C to 100 °C using the formula:

q = m × Cs × ΔT

where q is the heat in Joules, m is the mass in grams, Cs is the specific heat capacity (4.184 J/g°C for water), and ΔT is the change in temperature.


For the first step:

q1 = (422 g) × (4.184 J/g°C) × (100°C - 23.5°C)


Then, calculate the heat required to vaporize the water at 100 °C using the heat of vaporization:

q2 = m × ΔHvap

where ΔHvap is the heat of vaporization (2260 J/g for water).

For the second step:

q2 = (422 g) × (2260 J/g)


Combine both amounts of heat:

total heat (qtotal) = q1 + q2

Calculate the totals for q1 and q2, and then sum them to find the total heat required.

User Dmitry Khamitov
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