A 5.0 g ice cube at −20°C is in a rigid, sealed container from...

Question

A 5.0 g ice cube at −20°C is in a rigid, sealed container from which all the air has been evacuated. How much heat is required to change this ice cube into steam at 200°C? Steam has c_V= 1500 J/kg K and c_P= 1960 J/kg K.

Accepted Answer

Hey, everyone. So this problem is dealing with heat transfer and phase changes. Let's see what it's asking us. We have a 28 g ice block placed inside of a vacuum container. The initial temperature of the block is negative 15 degrees Celsius. And we're asked to determine the amount of heat transferred to the ice cube for it to melt and turn into steam at 120 degrees Celsius. Our multiple choice answers here are a 24 kilojoules B 30 kilojoules C 36 kilojoules or D 77 kilojoules. So we actually have three different phase changes that are happening in this problem. So let's talk through each one. First. We have ice and that turns into water, then we have water that turns into steam. And although this isn't actually the third one is not actually a phase change because it's staying a um vapor or gass, we have steam that's going to superheated steam. So each one of these steps has a set of equations associated with it. So we'll call this Q one from ice to water. Q two will be from water to steam and Q three will be from sea steam to superheated steam. So we can recall that our um heat energy is given by the equation MC delta T. So that's going to be the mass of the water multiplied by the specific heat capacity of ice multiplied by the delta T. So that's going to be the t of when it turns to liquid or water minus the temperature of the ice. And then we also have are latent heat of fusion term because we are changing from a solid to a liquid. And so that's going to be the mass of the water multiplied by the latent heat of fusion of the water. Similarly, when we go from water to steam will have the mass of the water, the specific heat capacity of the liquid water. And then our delta T term is going to be the temperature of the steam when it initially turns to steam minus the temperature of the liquid. And then we have the mask and we add to that the mass of the water multiplied by the latent heat of vaporization of the water because we are vaporizing, going from water to steam. Then the last step from steam to superheated steam. Again, we don't actually have a phase change here. So it's just going to be the mass of the water, the specific heat capacity of steam. And then our delta T is going to be the temperature of the superheated steam minus that initial temperature of just the steam. And so when we look at this total heat transferred, we have our Q total is going to be equal to the sum of these. So Q one plus Q two plus Q three. And we can see that we have a mass of the water term in each of these terms. So we will factor that out. And so this equation becomes the mass of the water. Now, and the problem that was given to us as 25 g, we're going to rewrite that to keep us in standard units. That's gonna, that's going to be 0.25 kg multiplied by all of these other terms. OK. So one at a time, we have the specific heat capacity of ice, which we can recall is 2090 Jews per kilogram. Kelvin multiplied by hour temperature of the liquid. So we can recall the ice turns to liquid or water turns to liquid at zero degrees Celsius minus the initial temperature of the ice was negative 15 degrees Celsius. And so those negatives cancel plus our latent heat of fusion of water, which you can recall is 3.34 multiplied or excuse me, 3.34 times 10 to the fifth Jews per kilogram. And so that's our Q one term. So plus R Q two charm is going to be similarly 4190 jewels per kilogram. Kelvin for our specific heat capacity of liquid water multiplied by the temperature of steam, which we, we can recall is 100 degrees Celsius minus the initial temperature of water or when it becomes a liquid, which again, we've established was zero degrees Celsius plus our specific or uh the specific heat of fusion of vapor of vaporization of water, excuse me, which is 2.26 times 10 to the six jewels per kilogram. And so that's our Q two term. And then for Q three, we simply have our specific heat capacity of steam, which you, you can recall is 1996 Jews per kilogram. K multiplied by our delta T. So the problem tells us we're taking this steam up to 120 degrees Celsius and that's from an initial steam temperature of 100 degrees Celsius. And so when we plug that into our calculator, we get a total heat energy of 7. times 10 to the fourth jules. In one last step, the answer was in kills. So that becomes 77 S and when we look at our multiple choice answers, it aligns with answer choice D. So D is the correct answer for this problem. That's all we have for this one. We'll see you in the next video.

A 5.0 g ice cube at −20°C is in a rigid, sealed container from which all the air has been evacuated. How much heat is required to change this ice cube into steam at 200°C? Steam has c_V= 1500 J/kg K and c_P= 1960 J/kg K.

Key Concepts

Phase Changes

Specific Heat Capacity

Latent Heat

Watch next

A 5.0 g ice cube at −20°C is in a rigid, sealed container from which all the air has been evacuated. How much heat is required to change this ice cube into steam at 200°C? Steam has cV = 1500 J/kg K and cP = 1960 J/kg K.

Key Concepts

Phase Changes

Specific Heat Capacity

Latent Heat

Watch next

A 5.0 g ice cube at −20°C is in a rigid, sealed container from which all the air has been evacuated. How much heat is required to change this ice cube into steam at 200°C? Steam has c_V= 1500 J/kg K and c_P= 1960 J/kg K.