Question

how much energy (in kj) is required to heat 25.0 g of H2O (s) at -18.0 degrees celsius to H2O (g) at 130.0 degrees celsius

Answer 1

H₂O at -18 degrees is solid. When we heat it till 0°C, it melts and becomes liquid. After 100 °C, it evaporates and becomes gaseous. Therefore, the process of heating of solid H₂O at -18 °C to gaseous H₂O at 130°C must be separated in 5 different stages:

1) heating of solid H₂O up to 0°C: $Q = c_s·m·∆T$ , where $c_s$ is the specific heat capacity of solid water and $∆T$ is the difference temperature: 0-(-18) = 18. Note that the difference temperature does not need to be converted in kelvin.

2) melting of solid H₂O: $Q = ∆H_{Fus}·m$ , where $∆H_{Fus}$ is the specific heat of fusion of water, 333.55 J/g;

3) heating of liquid H₂O to 100°C: $Q = c_l·m·∆T$ , where $c_l$ is the specific heat capacity of liquid water, and $∆T$ is 100-0, or 100;

4) vaporisation of water: $Q = ∆H_{Vap}·m$ , where $∆H_{Vap}$ is the specific heat of vaporisation of H₂O, 2257 J/g;

5) heating of gaseous H₂O: $Q = c_g·m·∆T$ , where $c_g$ is the specific heat capacity of gaseous water and $∆T$ is 130-100, or 30.

Let's neglect the dependence of the specific heat capacity on temperature for simplicity. The specific heats of solid, liquid, and gaseous water are 2.05, 4.1813 and 2.080 J/g/K, respectively.

Therefore:

$E = ∑Q$

$E= (2.05·18 + 333.55 +4.1813·100+2257+2.080·30)·25.0$

$E = 77.7$ kJ

Answer: 77.7 kJ is required to heat 25.0 g of H₂O(s) at -18.0 °C to H₂O(g) at 130.0 °C.