Question

If 25 kg of ice at 0°C is combined with 4 kg of steam at 100°C, what will be the final equilibrium temperature (in °C) of the system?

Answer 1

Answer:

The temperature of the final equilibrium temperature of the system is 19.7⁰C

Explanation:

Given;

mass of the ice, m₁ = 25 kg

temperature of the ice =  0°C

mass of the steam, m₂ = 4 kg

temperature of the steam, = 100 ⁰C

Let the temperature of the resulting mixture = t

Apply the principle of conservation of energy.

Heat required to melt the ice + Heat gained by the mixture = Heat required to convert the water to steam + Heat lost by the mixture

$m_1L_f + m_1c (t - 0) = m_2L_v + m_2c(100 - t)$

where;

Lf is the latent heat of fusion of ice =  3.33 x 10⁵ J/kg.

Lv is the latent heat of vaporization of water, = 2.260 x 10⁶ J/kg

c is the specific heat capacity of water = 4,200 J/kg

$m_1L_f + m_1c (t - 0) = m_2L_v + m_2c(100 - t)\\\\25 \times 3.33\times 10^5 + 25\times 4,200 (t ) = 4\times 2.26\times 10^6 + 4\times 4,200(100 - t)\\\\8,325,000 + 105,000t = 9,040,000 + 1,680,000 - 16,800t\\\\105,000t + 16,800t = 10,720,000 - 8,325,000\\\\121,800t = 2,395,000\\\\t = \frac{2,395,000}{121,800} \\\\t = 19.7 \ ^0 C$

The temperature of the final equilibrium temperature of the system is 19.7⁰C