There is a problem that asks if an ice cube is dropped into nitrogen (boiling point of nitrogen is 77K), how much nitrogen will evaporate?
heat lost by ice = heat gained by nitrogen
But by is the equation
m(ice)*c(ice)*deltaT = m(nitrogen)*L(vaporation) ?
Why isn't it
m(ice)*c(ice)*deltaT + m(ice)*L(ice) = m(nitrogen)*L(vaporation)
because aren't you melting the ice into water too?
thanks for the help.
You're correct in considering the heat required for both phase changes (solid to liquid and liquid to gas) when analyzing the situation. Let's break down the heat transfer equations in each case:
1. Melting of ice:
The equation should be:
m(ice) * c(ice) * ΔT + m(ice) * L(fusion) = m(water) * c(water) * ΔT
Where:
m(ice) = mass of ice
c(ice) = specific heat capacity of ice
ΔT = change in temperature (final temperature - initial temperature)
L(fusion) = latent heat of fusion (heat required for solid to liquid phase change)
m(water) = mass of water formed from melting the ice
c(water) = specific heat capacity of water
2. Vaporization of water into nitrogen:
The equation should be:
m(water) * c(water) * ΔT + m(water) * L(vaporization) = m(nitrogen) * c(nitrogen) * ΔT
Where:
m(water) = mass of water
L(vaporization) = latent heat of vaporization (heat required for liquid to gas phase change)
m(nitrogen) = mass of nitrogen
c(nitrogen) = specific heat capacity of nitrogen
To sum up, you have to consider the heat required for both processes: melting the ice and vaporizing the water. Therefore, the correct equation should include both terms:
m(ice) * c(ice) * ΔT + m(ice) * L(fusion) = m(nitrogen) * L(vaporization)
To properly assess the situation, let's break down the heat transfers happening in this problem.
1. Heat lost by the ice cube:
The ice is initially at a higher temperature, likely at 273.15 K (the melting point of ice) or lower. However, we don't have the exact initial temperature, so let's assume it is at the melting point.
The heat lost by the ice can be calculated using the formula:
Q1 = m(ice) * c(ice) * ΔT
where:
- Q1 is the heat lost by the ice cube
- m(ice) is the mass of the ice cube
- c(ice) is the specific heat capacity of ice (approximately 2.09 J/g·K)
- ΔT is the change in temperature from the initial temperature of the ice to the final temperature (77 K).
2. Heat gained by the nitrogen:
The nitrogen is initially at its boiling point, 77 K. When the ice cube is dropped into the nitrogen, it will cause some of the nitrogen to evaporate into gas. So, the heat gained by the nitrogen can be calculated using the equation:
Q2 = m(nitrogen) * L(vaporization)
where:
- Q2 is the heat gained by the nitrogen
- m(nitrogen) is the mass of the nitrogen evaporated
- L(vaporization) is the latent heat of vaporization of nitrogen (approximately 199.1 J/g)
Now, we need to consider the additional heat required to melt the ice into water. This is given by:
Q3 = m(ice) * L(fusion)
where:
- Q3 is the heat required to melt the ice into water
- L(fusion) is the latent heat of fusion of ice (approximately 333.55 J/g)
However, in this specific problem, we are not asked to consider the heat required to melt the ice into water. The question only asks how much nitrogen will evaporate. Therefore, we only need to calculate Q1 and Q2.
To summarize, the correct equation for this problem is:
Q1 = m(ice) * c(ice) * ΔT
Q2 = m(nitrogen) * L(vaporization)
Since the heat lost by the ice is equal to the heat gained by the nitrogen:
Q1 = Q2
m(ice) * c(ice) * ΔT = m(nitrogen) * L(vaporization)
I hope this clears up any confusion you had.