How much heat is required to convert 2.73 g g of water at 32.0 ∘ C ∘C to steam at 100.0 ∘ C ∘C ? (Δ H vap ( H 2 O)=40.7kJ/mol, Heatcapacity( H 2 O)=4.184J/g ∘ C) (ΔHvap(H2O)=40.7kJ/mol, Heatcapacity(H2O)=4.184J/g∘C) Express your answer with the appropriate units.

We need 6942.7 J of heat

Explanation:

Step 1: Data given

Mass of water = 2.73 grams

Initial temperature of water = 32.0 °C

Final temperature of water = 100 °C

Heat capacity of water = 4.184 J/g°C

ΔHvap(H2O) = 40.7kJ/mol

Step 2 : Calculate heat required to heat water from 32.0 °C to 100 °C

Q = m*c*ΔT

⇒with Q = the heat required = TO BE DETERMINED

⇒with m= the mass of water = 2.73 grams

⇒with c = the specific heat of water = 4.184 J/g°C

⇒with ΔT = The change of temperature = T2 - T1 = 100 °C - 32. 0°C = 68.0 °C

Q = 2.73 * 4.184 * 68.0 °C

Q = 776.7 J

Step 3: Calculate moles H2O

Moles H2O = mass H2O / molar mass H2O

Moles H2O = 2.73 grams / 18.02 g/mol

Moles H2O = 0.1515 moles

Step 4: Calculate heat required to change water from liquid to gas phase

Q = moles H2O*ΔHvap(H2O)

Q = 0.1515 moles * 40700J/mol

Q = 6166 J

Step 5 : Calculate the total heat required

Q = 776.7 J + 6166 J

Q = 6942.7 J

We need 6942.7 J of heat

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