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Suppose that 0.82 g of water condenses on a 75.0 g block of iron that is initial

ID: 478385 • Letter: S

Question

Suppose that 0.82 g of water condenses on a 75.0 g block of iron that is initially at 22 degrees Celsius. If the heat released during condensation goes only to warming the iron block, what is the final temperature (in degrees Celsius) of the iron block? (Assume a constant enthalpy of vaporization for water of 44.0 kJ/Molly). Suppose that 0.82 g of water condenses on a 75.0 g block of iron that is initially at 22 degrees Celsius. If the heat released during condensation goes only to warming the iron block, what is the final temperature (in degrees Celsius) of the iron block? (Assume a constant enthalpy of vaporization for water of 44.0 kJ/Molly). If the heat released during condensation goes only to warming the iron block, what is the final temperature (in degrees Celsius) of the iron block? (Assume a constant enthalpy of vaporization for water of 44.0 kJ/Molly).

Explanation / Answer

Given that;

Mass of water = 0.82 g

Mass of iron block = 75.0 g

Initial temperature = 22 degree

vaporization for water of 44.0 kJ/Molly

Moles of water = mass of water/molar mass of water

= 0.82/18.02 = 0.0455 mol

Heat released by water condensing

= moles of water x enthalpy of vaporization

= 0.0455 x 44.0 x 103 = 2002.2 J

Let T be the final temperature of the iron block

Heat absorbed by iron block

= mass of iron block x specific heat of iron x temperature change

= 75.0 x 0.450 x (T - 22)

heat absorbed = heat released

75.0 x 0.450 x (T - 22) = 2002.5

33.75 T – 742.5 = 2002.5

33.75 T = 742.5 + 2002.5

33.75 T =2745

T = 81.3 oC

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