Question

1. At 1 atm, how much energy is required to heat 91.0 g of H2O(s) at –24.0 °C to H2O(g) at 171.0 °C?

2.When 1518 J of heat energy is added to 48.6 g of hexane, C6H14, the temperature increases by 13.8 °C. Calculate the molar heat capacity of C6H14.

3.A 56.90 g sample of a substance is initially at 21.3 °C. After absorbing 1615 J of heat, the temperature of the substance is 146.8 °C. What is the specific heat (c) of the substance?

Answer 1

1.

Q = Q1 + Q2 + Q3 + Q4 + Q5

Q1 = heat required to get ice at 0 degreeC from - 24 degreeC

Q2 = heat required to melt ice to water at 0 degreeC

Q3 = heat required to increase temperature of water from 0 degreeC to 100 degreeC

Q4 = heat required to change water to vapor at 100 degreeC

Q5 = heat required to increase temperature of vapor to 171 degreeC

Now,

Q1 = m c * dT = 91 g * 2.108 J/g-degreeC * 24 degreeC = 4.603 kJ

Q2 = mLf = 91 g * 334 J/g = 30.394 kJ

Q3 = m * c * dT = 91 g * 4.184 J/g-degreeC * 100 degreeC = 38.074 kJ

Q4 = m* Lv = 91 g * 2259 J/g = 205.569 kJ

Q5 = m * c * dT = 91 g * 2.02 J/g-degreeC * 71 degreeC = 13.051 kJ

Total heat = 4.603 + 30.394 + 38.074 + 205.569 + 13.051 = 291.151 kJ

2.

Q = n * c * dT

1518 J = 48.6 g / 86 g/mol * c * 13.8 degreeC

c = 1518 * 86 / ( 48.6 * 13.8) J/mol-degreeC

c = 194.65 J / mol-degreeC

3.

Q = m * c * dT

1615 J = 56.9 g * c * ( 146.8 - 21.3) degreeC

1615 J = 56.9 g * c * 125.5 degreeC

c = 1615 / ( 56.9 * 125.5) J/g-degreeC

c = 0.226 J/g-degreeC