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

conduction is the transfer of thermal energy (heat in transfer) due to collisions between the molecules in the object. collisions between adjacent atoms and molecules transfer kinetic energy from the warmer to the cooler object. the objects must be in physical contact.

5.6⋅10−^19  j

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