Answer:
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
