Answer to Question #80820 in General Chemistry for cade

Question

At 1 atm, how much energy is required to heat 45.0 g of H2O(s) at –20.0 °C to H2O(g) at 129.0 °C? Helpful constants can be found here

Answer 1

There are five things to calculate here:
--the energy needed to raise the ice to 0°C
--the energy needed to melt the ice
--the energy needed to raise the water to 100 °C
--the energy needed to boil the water
--the energy needed to raise the steam to 133 °C.
The specific heat capacity of ice is about 2.03 J/(g·K).
The specific enthalpy of fusion of water is 333.55 J/g at 0 °C.
The specific heat capacity of water is 4.1927 J/(g·K) (average).
The heat of vaporization of water at 100 °C is 2256.2 J/g.
The heat capacity of steam at 100 °C is about 2.080 J/(g·K).
To heat the ice, we need
(45.0 g)(2.03 J/(g·K))(20.0 K) = 1827 J.
To melt the ice, we need
(45.0 g)(333.55 J/g) = 15010 J.
To heat the water, we need
(45.0 g)(4.1927 J/(g·K))(100 K) = 18870 J.
To boil the water, we need
(45.0 g)(2256.2 J/g) = 101530 J.
To heat the steam, we need
(45.0 g)(2.080 J/(g·K))(29.0 K) = 2714.4 J.
Total = 139951.4 J.