A 4.00-kg silver ingot is taken from a furnace, where its tempera...

Question

A 4.00-kg silver ingot is taken from a furnace, where its temperature is 750.0°C, and placed on a large block of ice at 0.0°C. Assuming that all the heat given up by the silver is used to melt the ice, how much ice is melted?

Accepted Answer

Hey everyone in this problem. A one kg iron fajitas sizzling plate with a temperature of degrees Celsius is removed from the oven, a big ice cube at zero degrees Celsius is placed on the plate and were asked how much ice melts. If we assume that all the heat released by the iron plate is transferred to the block of ice. Alright, now, if all the heat released by the iron plate is transferred to the block of ice, that tells us the quantity of heat Q. For the system, it's going to equal to zero. Okay, We have no net heat lost or gained because it's all being transferred. Now, what makes up the quantity of heat for the system? No, well, we have two things. Okay, we have quantity of heat Q. For the ice. The quantity of heat Q. For the sizzling point. All right now, we're looking to find how much ice melts. When we're talking about how much ice melts, we're looking for the mass of ice. Okay? And if we want the massive ice will recall that that's going to be related to Q. Okay, so let's use this equation and see if we can figure out the massive ice. Okay, So let's start with Q. The quantity of heat for ice. Okay, what's going to happen to ice? Well, the ice is going to melt. Okay, the ice is sitting at 0°C. It's being placed on that hot plate. Okay, so the temperature is going to rise and we know that it's going to melt at 0°C. Okay, so it's going to melt. So this is a phase change. All right, now, when we're talking about quantity of heat Q. For a phase change, this is going to be equal to M. Okay. And in this case the massive ice times the latent heat of fusion. Okay. Because we're going from um a solid to a liquid. We have the latent heat of fusion F. L. F. Sorry? Okay. And this is positive because we have heat entering into our material. The ice is warming up to melt into water with heat entering and so this is positive. All right, well, this is going to be equal to m ice. We don't know the massive ice. This is actually what we're trying to find many times the latent heat of fusion. Okay, for ice or water, it's 334 Times 10 to the three jewels per kilogram. Okay. Alright. So we figured out everything we can for the quantity of heat for the ice. Now, let's do that for the plate. The plate. What's going to happen to the plate? Okay, well, this is iron. It's not going to change. It's not going to go through a face change. It's not going to melt or boil. Okay, But it is going to cool. Okay, you're adding an ice cube onto it. That's gonna cool down that plate. Okay. And so we have a temperature change. Well, we write out Q. The quantity of heat for a temperature change. This is going to be equal to M. C. Times DELTA T. Okay and in this case we have the mass we'll call mp for the plate. Okay? C. Times delta T. Okay. We're looking at sea the specific heat capacity. Okay. And in this case our plate is iron. So we're gonna want the specific heat capacity for iron. Alright, so let's work this out. The mass of the plate is one kg. See the specific heat capacity for iron. You can look this up in the table in your textbook or that your professor provided. It's gonna be 470 jewels per kilogram Calvin times delta T. Alright, now the final temperature Going to be zero. Okay. We're Gonna Add 273.15 to convert this to Calvin. And we know that our initial temperature is 250 degrees Celsius. Okay so we're going to subtract 250 degrees Celsius plus 273.15. And let me get rid of this degree Celsius here um because this converts to Calvin. Okay. Alright great. So if we work this out we get the quantity of heat for the plate is going to be negative 117,500 jewels. Alright so we know quantity of heat Q. For the plate we know the quantity of heat que for the ice. Okay. Which is a function of the mass of the ice. Which is what we're looking for. Let's put this all together again. We know that the quantity of heat for the system is going to equal zero. Okay. And it's gonna be the sum of Q. Ice plus cute plate. So we have zero is equal to The mass of the ice. Um ice times 334 times 10 to the three jewels per kilogram Plus quantity of heat for the plate which is negative 117,500 jewels. Moving that to the other side, we get the massive ice and ice is going to be equal to 117,500 jewels divided by 334 times 10 to the three jewels per kilogram. Okay, when we divide jewel, sorry, jewel divided by jewel per kilogram is going to leave us with the unit of kg which is a unit of mass that we want. Okay, we get 0. kg. Alright. And if we look at our answer choices, okay? We see that they're in grams. So we want to convert to grams in order to convert kilograms into grams, recall that we multiply by 1000 and this is going to give us 352 g. We go up. We see that we have option B Okay The amount of ice that's going to melt is going to be B- 352 g. Thanks everyone for watching. I hope this video helped see you in the next one.

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A 4.00-kg silver ingot is taken from a furnace, where its temperature is 750.0°C, and placed on a large block of ice at 0.0°C. Assuming that all the heat given up by the silver is used to melt the ice, how much ice is melted?

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