Question

1 kg of water (specific heat = 4184 J/(kg K)) is heated from freezing (0°C) to boiling (100°C). What is the change in thermal energy?

Answer 1

Answer: 1560632 joules

Step-by-step explanation:

The change in thermal energy (Q) required to heat ice depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)

Thus, Q = MCΦ

Given that:

Q = ?

Mass of frozen water (ice) = 1kg

C = 4184 J/(kg K)

Φ = (Final temperature - Initial temperature)

= 100°C - 0°C = 100°C

Convert 100°C to Kelvin

(100°C + 273) = 373K

Then, Q = MCΦ

Q = 1kg x 4184 J/(kg K) x 373K

Q = 1560632 joules

Thus, the change in thermal energy is 1560632 joules