For a class demonstration, your physics instructor pours 1.16 kg of steam at 100.0°C over 4.69 kg of ice at 0.0°C and allows the system to reach equilibrium. He is then going to measure the temperature of the system. While the system reaches equilibrium, you are given the latent heats of ice and steam and the specific heat of water: Lice = 3.33·105 J/kg, Lsteam = 2.26·106 J/kg, cwater = 4186 J/(kg °C). You are asked to calculate the final equilibrium temperature (in °C) of the system. What value do you find?
It is a simple technique: add the heats gained, and set that sum to zero.
HeatIcemelting+heatwater+heatsteamcondnesing + heatgained steam cooling=0
massice*Hc+ massice*cwater(Tf-0)+masssteam*Hv +masssteam*cw*(Tf-100)=0
solve for Tf.
To calculate the final equilibrium temperature of the system, we can use the principle of energy conservation.
First, we need to determine the amount of heat transferred between the steam and ice until they reach equilibrium.
Let's calculate the heat transferred from the steam to melt the ice:
Q1 = mass of ice * latent heat of fusion
Q1 = 4.69 kg * 3.33 * 10^5 J/kg
Q1 = 1.55917 * 10^6 J
Next, let's calculate the heat transferred from the steam to raise the temperature of the melted ice to the final equilibrium temperature:
Q2 = mass of water * specific heat of water * change in temperature
Q2 = 4.69 kg * 4186 J/(kg °C) * (final temperature - 0°C)
Q2 = 1.959634 * 10^4 * (final temperature - 0°C)
Finally, let's calculate the heat transferred from the steam to vaporize the melted ice:
Q3 = mass of water * latent heat of vaporization
Q3 = 4.69 kg * 2.26 * 10^6 J/kg
Q3 = 1.05894 * 10^7 J
Since the system reaches equilibrium, the heat transferred from the steam to the ice is equal to the sum of Q1, Q2, and Q3:
Q1 + Q2 + Q3 = 1.55917 * 10^6 J + 1.959634 * 10^4 * (final temperature - 0°C) + 1.05894 * 10^7 J = 0
Now, we can solve this equation for the final equilibrium temperature (in °C).