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What amount of heat, in kJ, is required to convert 2.90 g of water at 67.0 °C to 2.90 g of steam at 100.0 °C? (specific heat capacity of water = 4.184 J/g • °C; ∆Hvap = 40.7 kJ/mol)

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Answer:The heat required to convert a given mass of a substance from one phase to another can be calculated using the following formula:

q = m × ΔH

where:

q = heat (in joules or kilojoules)

m = mass of the substance (in grams or kilograms)

ΔH = enthalpy change (in J/g or kJ/mol) associated with the phase transition

In this case, we need to calculate the heat required to convert 2.90 g of water at 67.0 °C to steam at 100.0 °C. The specific heat capacity of water is 4.184 J/g • °C, and the enthalpy of vaporization (ΔHvap) of water is 40.7 kJ/mol.

First, we need to calculate the heat required to raise the temperature of the water from 67.0 °C to its boiling point of 100.0 °C:

q1 = m × c × ΔT1

q1 = 2.90 g × 4.184 J/g • °C × (100.0 °C - 67.0 °C)

q1 = 2.90 g × 4.184 J/g • °C × 33.0 °C

q1 = 4591.632 J

Next, we need to calculate the heat required to vaporize the water at its boiling point:

q2 = n × ΔHvap

q2 = (m/M) × ΔHvap

q2 = (2.90 g / 18.015 g/mol) × 40.7 kJ/mol

q2 = 0.1619 kJ

Finally, we can add the two heat values to obtain the total heat required:

q = q1 + q2

q = 4591.632 J + 0.1619 kJ

q = 4.7521 kJ

So, the amount of heat required to convert 2.90 g of water at 67.0 °C to 2.90 g of steam at 100.0 °C is approximately 4.7521 kJ.

Step-by-step explanation:

User Christian Schmitt
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