The molar heat capacity of mercury is ymol . K. What is the specihic heat capaci

Question

The molar heat capacity of mercury is ymol . K. What is the specihic heat capacity of this metal in , (4) A 182-g sample of gold at some temperature was added to 32.1 g of water. The initial water temperature was 25.0 °C. and the final temperature was 27.5 "C. If the specific heat capacity of gold is 0.128 ys-K, what was the initial temperature of the gold sample? tps Ethanol, CH,OH, boils at 78.29 "C. How much energy, in joules, is required to raise the temperature of 1.00 kg of ethanol from 20.0 °C to the boiling point and then to change the liquid to vapor at that remperature? (The specific heat capacity of liquid ethanol is 2.44 g K, and its enthalpy of vaporization i )t) , Adding 5.44 g of NH.NO (s) to 150.0 g of water in a coffee-cup calorimeter (with stirring to dissolve the salt) resulted in a decrease in temperature from 18.6 C to 16.2 Calculate the enthalpy change for dissolving NH,NO,(s) in water, in kJImol. Assume the solurion (whose mass is 155. g) has a specific heat capacity of

Explanation / Answer

1.

Cp = 28.1 J/mol.K

molar mass of Hg = 200.59 g/mol

1 mol Hg = 200.59g

replace 1 mol = 200.59g

28.1 J/200.59g.K = 0.14 J/g.K

2.

heat lost by gold = heat gain by water

final temperature should be equilibrium temperature ie. final temperature of gold and water should be same

heat lost = mgCpgT

T = change in gold temperature

heat gain by water = mwCpT'

T' = change in water temperature

182g x 0.128J/g.K x (T-27.5) = 22.1g x 4.18J/g.K x (27.5 - 25)

T = 37.41 oC

note 1K change = 1oC change

3.

first we have to heat ethanol from 20 oC to 78.29 oC in liquid state

then vaporization of ethanol takes place at that temperature

1kg = 1000g

sensible heat = mCpT = 1000g x 2.44J/g.K x (78.29 oC - 20 oC) = 142227.6 J

now heat required to vaporize = 1000g x 855J/g = 855000 J

total heat needed = 855000J + 142227.6J = 997.23 kJ

please ask remaining question seperately

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