Question

At 1 bar, how much energy is required to heat 69.0 g of H2O(s) at –24.0 °C to H2O(g) at 163.0 °C?

STRATEGY: 1. Calculate the energy needed for each temperature change or phase change individually. a. the energy needed to heat 69.0 g of H2O(s) from –24.0 °C to its melting point. b. the energy needed to melt 69.0 g of H2O(s) at its melting point. c. the energy needed to heat 69.0 g of H2O(l) from the melting point to the boiling point. d. the energy needed to boil 69.0 g of H2O(l) at its boiling point. e. the energy needed to heat 69.0 g of H2O(g) from the boiling point to 163.0 °C.

2. Sum the energies from each step and convert to kilojoules.

Step 1:

a.) Going from –24.0 °C to the melting point of H2O (0 °C) is a temperature change of 24.0 °C. How much energy is needed to heat 69.0 g of H2O(s) by 24.0 °C if its specific heat is 2.087 J/(g·°C)?

b. How much energy is needed to melt 69.0 g of H2O(s) if its enthalpy of fusion is 333.6 J/g?

c.) For H2O, going from the melting point (0 °C) to the boiling point (100 °C) is a temperature change of 100 °C. How much energy is needed to heat 69.0 g of H2O(l) by 100 °C if its specific heat is 4.184 J/(g·°C)?

d. How much energy is needed to boil 69.0 g of H2O(l) if its enthalpy of vaporization is 2257 J/g?

e.) Going from the boling point of H2O (100 °C) to 163.0 °C is a temperature change of 63.0 °C. How much energy is needed to heat 69.0 g of H2O(g) by 63.0 °C if its specific heat is 2.000 J/(g·°C)?

Answer 1