Question

At 1 atm, how much energy is required to heat 93.0 g of H2O(s) at –16.0 °C to H2O(g) at 153.0 °C?

Answer 1

1. The energy needed to raise the ice to 0°C :

Here we use The specific heat capacity of ice at −10 °C is about 2.05 J/(g·K).

To heat the ice:
(93.0 g)(2.05 J/(g·K))(16.0 K) = 3050.4 J.

2. The energy needed to melt this ice:

Here we use the specific enthalpy of fusion of water is 333.55 J/g at 0 °C.

(93.0 g)(333.55 J/g) = 31020.15 J.

3. The energy needed to raise the water upto 100 °C :

Here we use the specific heat capacity of water is 4.1927 J/(g·K)

(93.0 g)(4.1927 J/(g·K))(100 K) = 38992.11 J.

4. The energy needed to boil the water:

Here we use the heat of vaporization of water at 100 °C is 2256.2 J/g

(93.0 g)(2256.2 J/g) = 209826.6 J.

5. the energy needed to raise the steam to 153 °C.

Here we use the heat capacity of steam at 100 °C is about 2.080 J/(g·K).

(93.0 g)(2.080 J/(g·K))(53.0 K) = 10252.32 J.

Total =( 3050.4 +31020.15 +38992.11 + 209826.6 +10252.32 )J

=293141.58 J or 293.14158 KJ