At 1 atm, how much energy is required to heat 87.0 g of H2O(s) at –22.0 °C to H2O(g) at 147.0 °C?
now
heat required to bring Ice at -22 C to ice at O C
Q1 = m x s x dT
Q1 = 87 x 2.06 x 22
Q1 = 3942.84
now
heat required to melt ice at O c to water at O C
Q2 = m x dHf
Q2 = 87 x 334.16
Q2 = 29071.92
now
heat required to bring water at O C to water at 100 C
Q3 = m x s xdT
Q3 = 87 x 4.184 x 100
Q3 = 36400.8
now
heat required to convert water at 100 C to steam at 100 C
Q4 = m x dHvap
Q4 = 87 x 2259
Q4 = 196533
now
heat required to bring steam at 100 C to steam at 147 C
Q5 = m x s x dT
Q5 = 87 x 2.02 x 47
Q5 = 8259.78
now
total heat = Q1 + Q2 + Q3 + Q4 + Q5
so
total heat = 3942.84 + 29071.92 + 36400.8 + 196533 + 8259.78
total heat = 274208 J
total heat = 274.208 kJ
so
274.208 kJ of energy is required
Get Answers For Free
Most questions answered within 1 hours.