Question

3.75 × 10⁵ calories of heat is given out by 5 kg of water at 100°C. Calculate the temperature of the cooled water. The specific heat capacity of water is 1000 calkg⁻¹°C⁻¹.

Answer 1

Given: m = 5 kg, c = 1000 calkg⁻¹°C⁻¹, Q = 3.75 × 10⁵

Heat lost by water (Q) = mcΔT

$$\Delta T = \frac{Q}{mc}$$

$$= \frac{3.75 \times 10^5}{5 \times 1000}$$

= 75°C

Since water cools, final temperature = 100°C − 75°C = 25°C