Suppose that 0.94 g of water condenses on a 75.0g block of iron that is initiall

Question

Suppose that 0.94 g of water condenses on a 75.0g block of iron that is initially at 21 C.

Part A

If the heat released during condensation goes only to warming the iron block, what is the final temperature (in C) of the iron block? (Assume a constant enthalpy of vaporization for water of 44.0 kJ/mol.)

Suppose that 0.94 g of water condenses on a 75.0g block of iron that is initially at 21 C.

Part A

If the heat released during condensation goes only to warming the iron block, what is the final temperature (in C) of the iron block? (Assume a constant enthalpy of vaporization for water of 44.0 kJ/mol.)

Explanation / Answer

Lv = 44.0 KJ/mol = 44.0 Kj/molx(1000J / 1KJ) = 44000 J/mol

Given the mass of water vapor, m = 0.94 g

Hence moles of water vapor, n = mass / molecular mass of water = 0.94 g / 18gx mol-1  = 0.052 mol

Let the final temperature of the iron block and water be T DegC

Initial temperature of water vapor = 100 DegC

Heat released during condensation only goes to heat the iron block

Now heat lost by water vapor = nxLv

= 0.052 mol x (44000 KJ /mol)

= 2288 J

s(iron) = 0.450(J/gxDegC)

Heat gained by iron block = mxs(iron)xdT = 75.0g x 0.450(J/gxDegC) x (T - 21)

= 33.75x(T-21)

Now applying the formulae

heat lost by water vapor = Heat gained by iron block

=>2288J =  33.75x(T-21)

=> T - 21 = 68 DegC

=> T = 68+21 = 89DegC (final temperature)

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