Question

A liquid X of specific heat capacity 1050 Jkg⁻¹K⁻¹ and at 90°C is mixed with a liquid Y of specific heat capacity 2362.5 Jkg⁻¹K⁻¹ and at 20°C, when the final temperature recorded is 50°C. Find in what proportion the weights of the liquids are mixed.

Answer 1

Let the masses of liquid X and liquid Y be m_X and m_Y, respectively.

Heat lost by X:

m_X × 1050 × (90 − 50)

= m_X × 1050 × 40

Heat gained by Y:

m_Y × 2362.5 × (50 − 20)

= m_Y × 2362.5 × 30

By the principle of the method of mixtures,

Heat lost = Heat gained

m_X × 1050 × 40 = m_Y × 2362.5 × 30

$$\frac{m_X}{m_Y} = \frac{2362.5 \times 30}{1050 \times 40}$$

$$= \frac{70875}{42000}$$

$$= \frac{27}{16}$$

∴ X : Y = 27 : 16