# Problem: Suppose 1.25 g of steam at 100°C condenses on an insulted 65.0 g block of iron (heat capacity 0.449 J/g°C) that is initially at 20.0°C The final temperature of the block is 85.0°C. Find the average heat of vaporization of water during the condensation. The heat capacity of water is 4.18 J/g°C.A. 26.2 kJ/mol B. 27.3 kJ/mol C. 44.0 kJ/mol D. -27.3 kJ/mol E. 34.4 kJ/mol

###### FREE Expert Solution

We’re being asked to determine the average heat of vaporization of water.

We will use the heat absorbed by the metal piece to calculate. Recall that heat can be calculated using the following equation:

$\overline{){\mathbf{q}}{\mathbf{=}}{\mathbf{mc}}{\mathbf{∆}}{\mathbf{T}}}$

q = heat, J

+qabsorbs heat
–qloses heat

m = mass (g)
c = specific heat capacity = J/(g·°C)
ΔT = Tf – Ti = (°C)

Notice that the sample steam will condense to liquid at 100°but the liquid will also eventually decrease in temperature as it reaches thermal equilibrium with the water present in the container:

heat will be released when steam condenses (phase change). Calculate heat (q) using the equation:

$\overline{){{\mathbf{q}}}_{{\mathbf{1}}}{\mathbf{=}}{\mathbf{m}}{\mathbf{·}}{\mathbf{∆}}{{\mathbf{H}}}_{{\mathbf{vap}}}}$

q = heat, J
m = mass (g)
ΔHvap = heat of vaporization, J/g

heat will be released when the temperature of the liquid decreases. Calculate heat (q) using the equation:

$\overline{){{\mathbf{q}}}_{{\mathbf{2}}}{\mathbf{=}}{\mathbf{mc}}{\mathbf{∆}}{\mathbf{T}}}$

total heat released will be the sum of the two

Based on the given system:

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###### Problem Details

Suppose 1.25 g of steam at 100°C condenses on an insulted 65.0 g block of iron (heat capacity 0.449 J/g°C) that is initially at 20.0°C The final temperature of the block is 85.0°C. Find the average heat of vaporization of water during the condensation. The heat capacity of water is 4.18 J/g°C.

A. 26.2 kJ/mol

B. 27.3 kJ/mol

C. 44.0 kJ/mol

D. -27.3 kJ/mol

E. 34.4 kJ/mol