Question

If 20 g steam initially at 1000C is added to 60 g of ice initially at 00C, then find the final equilibrium temperature of the mixture.

Answer 1

The heat energy lost by the steam is equal to the heat energy gained by the ice.

The expression for the loss or gain in heat energy during a phase change is

$Q=mL_f$

Here, m is mass, L_f is latent heat of fusion.

The expression for the loss or gain in heat energy during a phase change is

$Q=mL_v$

Here, m is mass, L_v is latent heat of vaporization.

The heat energy required to change the phase of ice at constant temperature 0 °C is

$Q_1=m_{ice}L_f$

The heat energy required to raise the temperature of water from 0 °C to equilibrium temperature t_e is

$Q_2=m_{ice}c_w(t_e – 0 \;°C)$

Here, m_ice is mass of ice and c_w is specific heat of water.

The heat energy required to change the phase of steam at constant temperature 100 °C is

$Q_3 = m_{steam}L_v$

Here, m_steam is mass of steam and L_v is latent heat of vaporization.

The heat energy required to lower the temperature of water from 100 °C to equilibrium temperature t_e is

$Q_4=m_{steam}c_w(100 \;°C -t_e)$

The heat energy lost by the steam is equal to the heat energy gained by the ice. Hence

Q_1+Q_2=Q_3+Q_4

Substitute the values of Q₁, Q₂, Q₃ and Q₄ in the above equation to solve for t_e.

$m_{ice}L_f + m_{ice}c_w(t_e – 0 \;°C) = m_{steam}L_v + m_{steam}c_w(100 \;°C -t_e) \\ m_{ice}c_wt_e+m_{steam}c_wt_e= m_{steam}L_v + m_{steam}c_w100\;°C -m_{ice}L_f \\ t_e(m_{ice}c_w + m_{steam}c_w) = m_{steam}L_v + m_{steam}c_w100\;°C -m_{ice}L_f \\ t_e = \frac{m_{steam}L_v + m_{steam}c_w100\;°C -m_{ice}L_f}{m_{ice}c_w + m_{steam}c_w} \\ m_{ice} = 60 \;g \\ m_{steam} =20 \;g \\ L_v=540 \;cal/g \\ c_w = 1.0 \;cal/g \cdot °C \\ L_f = 80 \;cal/g \\ t_e = \frac{20 \times 540 + 20 \times 1.0 \times 100 -60 \times 80 }{60 \times 1.0 + 20 \times 1.0} \\ = \frac{10800 + 2000 -4800}{80} \\ = 100 \;°C$

Therefore, the equilibrium temperature of the system is 100 °C.