Answer to Question #226435 in Molecular Physics | Thermodynamics for Jonathan Delelegn

Question #226435

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.


1
Expert's answer
2021-08-16T08:36:11-0400

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, Lf 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, Lv 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 te is

"Q_2=m_{ice}c_w(t_e \u2013 0 \\;\u00b0C)"

Here, mice is mass of ice and cw 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, msteam is mass of steam and Lv is latent heat of vaporization.

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

"Q_4=m_{steam}c_w(100 \\;\u00b0C -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 te.

"m_{ice}L_f + m_{ice}c_w(t_e \u2013 0 \\;\u00b0C) = m_{steam}L_v + m_{steam}c_w(100 \\;\u00b0C -t_e) \\\\\n\nm_{ice}c_wt_e+m_{steam}c_wt_e= m_{steam}L_v + m_{steam}c_w100\\;\u00b0C -m_{ice}L_f \\\\\n\nt_e(m_{ice}c_w + m_{steam}c_w) = m_{steam}L_v + m_{steam}c_w100\\;\u00b0C -m_{ice}L_f \\\\\n\nt_e = \\frac{m_{steam}L_v + m_{steam}c_w100\\;\u00b0C -m_{ice}L_f}{m_{ice}c_w + m_{steam}c_w} \\\\\n\nm_{ice} = 60 \\;g \\\\\n\nm_{steam} =20 \\;g \\\\\n\nL_v=540 \\;cal\/g \\\\\n\nc_w = 1.0 \\;cal\/g \\cdot \u00b0C \\\\\n\nL_f = 80 \\;cal\/g \\\\\n\nt_e = \\frac{20 \\times 540 + 20 \\times 1.0 \\times 100 -60 \\times 80 }{60 \\times 1.0 + 20 \\times 1.0} \\\\\n\n= \\frac{10800 + 2000 -4800}{80} \\\\\n\n= 100 \\;\u00b0C"

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


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS