How Much Steam Is Needed to Heat Water in a Copper Calorimeter to 50°C?

[Cummings]

A couple of anoying problems that i know i should be able to do but are just getting me frustrated.

1. A 50.0-g copper calorimeter contains 250g of water at 20 degrees C. How much steam must be condensed into the water if the final temperature of the system is to reach 50 degrees C

so, the water and the copper colorimeter are gaining heat and the steam is loosing heat via the latent heat of vaporisation.

so
.05 * 387 * (50-20) + .250 * 4180 * (50-20) = 2.26*10^6 * massofsteamrequired

580.5 + 31350 = 2.26*10^6 * massofsteamrequired
so mass of steam required = 14.12g

yet i am told it is meant to be 12.9g

2. In an insulated vessel, 250g of ice at 0 degrees C is added to 600 g of water at 18 degrees C
a, what is the final temperature of the system.
b, how much ice remains when the system reached equilibrium.

for a, the ice is at 0 degrees so it will gain energy from the water for the latent heat of fusion then gain energy as water to reach an equilibrium. the water will loose energy to reach an quilibrium.

i am stuck on this one. I am told the answer is 0 degrees C but i can't find how to get it.

 

[AKG]

For the second one, figure out how much heat is lost by that much water cooling down that much. Also, figure out how much heat is required to melt the ice. You will find that the heat required to melt the ice is greater than or equal to the heat lost in cooling the water. If they are equal, then the water will come down to zero, and the ice will just have finished melting. If the heat required to melt the ice is greater, then the ice will only be partially finished melting, and in fact the water will start to freeze.

For the first one, realize that the steam is at [itex]100^o\ C[/itex], so it will not only have to condense, but cool to [itex]50^o\ C[/itex]. That process itself will give off heat, so you won't need as much steam.

 

[M98Ranger]

I understand your frustration with these problems. It can be challenging to apply the concepts of specific and latent heat to specific scenarios. Let's break down each problem and see if we can find where the discrepancies may be coming from.

Problem 1: In this problem, we have a copper calorimeter containing 250g of water at 20 degrees C. We need to calculate how much steam needs to be condensed into the water in order for the final temperature of the system to be 50 degrees C.

You are correct in your approach of using the heat gained by the water and the copper calorimeter and the heat lost by the steam to find the mass of steam required. However, there seems to be a mistake in your calculation. The specific heat of water is 4180 J/kg*K, not 4180 J/g*K. So the correct calculation should be:

(0.05 kg * 387 J/kg*K * (50-20) K) + (0.25 kg * 4180 J/kg*K * (50-20) K) = (2.26*10^6 J/kg * mass of steam required)

580.5 J + 313500 J = 2.26*10^6 J/kg * mass of steam required

So, the mass of steam required is actually 12.9 g, not 14.12 g.

Problem 2: In this problem, we have an insulated vessel containing 250g of ice at 0 degrees C and 600g of water at 18 degrees C. We need to find the final temperature of the system and the amount of ice remaining when the system reaches equilibrium.

For part a, your approach is correct. The ice will first absorb energy from the water to melt (latent heat of fusion) and then absorb more energy to reach an equilibrium temperature. The water will also lose energy to reach an equilibrium temperature. So the equation should be:

(0.25 kg * 334000 J/kg) + (0.25 kg * 4180 J/kg*K * (Tf - 0) K) = (0.6 kg * 4180 J/kg*K * (Tf - 18) K)

83,500 J + 10450 J*Tf = 12,540 J*Tf - 225,720 J

So, the final temperature (Tf) of the system is 0 degrees C.

For part b