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A 5 kg piece of lead having a temperature of 80 â—¦C is added to500 g of water at a temperature of 20 â—¦C. What is the specific heatof the metal if the fin...

Question

A 5 kg piece of lead having a temperature of 80 â—¦C is added to500 g of water at a temperature of 20 â—¦C. What is the specific heatof the metal if the final equilibrium temperature of the system is35â—¦C?

A 5 kg piece of lead having a temperature of 80 â—¦C is added to 500 g of water at a temperature of 20 â—¦C. What is the specific heat of the metal if the final equilibrium temperature of the system is 35â—¦C?



Answers

A $2.50 \mathrm{~kg}$ lump of aluminum is heated to $92.0^{\circ} \mathrm{C}$ and then dropped into $8.00 \mathrm{~kg}$ of water at $5.00^{\circ} \mathrm{C}$. Assuming that the lump-water system is thermally isolated, what is the system's equilibrium temperature?

The energy conservation. We can write that the heat released by the copper. It will be gained by the water and L. A. Medium. So heat que corp er it will be equal to Q water plus Q aluminum. So substituting values so we get massive cooper a specific heat of copper and initial temperature of the copper minus final equilibrium temperature. City finals that will be equal to mass of water specific heat of water and final equilibrium temperature minus T. Initial of water plus mass of L. A. Medium specific heat of L. A. Medium and the temperature final minus temperature of aluminum. So from here after rearranging we will get temperature Final. This will be equal to mass of copper specific heat of cooper temperature of copper plus plus mass of water specific heat of water temperature of water plus massive L. A. Medium specific heat of L. A. Medium temperature of L. A. Medium divided by M W C W plus M L C A L. Plus and cooper C cooper. Okay, so substituting values as further questions. So the final it will be equal to massive cooper is too Who into massive copper is given us uh in the graham. Okay, so we can calculate, we can substitute all the data into graham. So two into 10 to the power to graham, manipulated by specific heat of copper is 0.386 closer per kilogram kelvin and manipulated by temperature of copper is 4 50 Calvin plus mass of water. It is It is 5-10 to the power to Graham and Specific heat of water. It will be 4.19. It will be 4.19 closer per kilogram kelvin And plus Manipulated by temperature of water. It is to 80 Calvin Plus massive aluminum. It is one into 10 to the power to Graham manipulated by specific heat of aluminum. It is 0.9 program program. Sorry, closer per program, Calvin and temperature is 200 Calvin Okay day whereby uh M W which is two and 2 10 to the power to and the specific heat of water, which is uh massive water is 5 to 10 to the power to. So 5 to 10 to the power to into 4.19 plus two into 10 to the power to into 0.386 plus one into 10 to the power to 10 to 0.9. So from here, after solving final equilibrium temperature, the final this will be equals 2 to 82.6 19 Calvin. Or it can be nearly taken as The final. It will be equals two 283 Calvin. Okay, so this will be the answer for the question. This is the final equilibrium temperature of the system. Okay,

This question we have metal and that is added to water. And we're told that there is a temperature change. The temperature of the water rises. No, he is lost to the surrounding. So we want the specific heat of the metal. So this is a cuticle EMC delta D. Problem. So as you know, the heat losses equal to he gained. So the metal is losing the heat. So we have negative M. C. Delta T. Metal is equal to positive M. C delta T. Water. The T final is going to be the same for both. They are both going to reach the same temperature. So therefore we substitute in the values and we leave see as a variable because that's a specific he of metal were looking for and then we will substitute in the rest of the values that we do know. So we solve for C. Mathematically and our value for C is going to be in jewels Program times degree C. Because it needs to have the same units as everything that was in our setup. So We end up with C is a good .487 jules. Program times degree Celsius. That means that The metal requires .487 jewels to heat one g of the medal by 1°.

We know that the heat capacity is the quantity off heat required to raise the temperature off a sample by one degree senses or one Kelvin is called us heat capacity. The formulation for heat capacity is represented over here. The calculate the amount off heat give to the water has shown below substituting the values into the formula. We have a mount off he transferred. That is 1.1 kill a jewel. The heat transferred between water and the metal is equal. Therefore, the heat transfer off metal is also the same. Now we can calculate the specific heat off the metal, substituting the heat transfer for the metal on dhe, the delta T that is changing temperature for the middle putting in the equation. First, that is this one. We can determine the specific heat off the metal that turns out to be zero point toe wear jewels by Graham. So

Hello. So we're gonna use the concept of energy conservation so that he'd lost by the copper will be called the heat gain by the water at room temperature. Um Oh so this is the expression for finding heat. The M. C. M. Is the mass, sees the specific heat capacity and then built A. T. Is the change in temperature for copper. And we have to something called water on the right. You notice in the question that we're not giving the mass of water, but it can be found was a mass, its density and volume. So this density of water is density in uh kill grandpa middle scoop and then the volume was given in leaders. So you convert that to miniscule. Uh he's going to give you this mask which is .499 Kg. So this is the mass of the water in the system. Uh So once you have that we just do this substitution for each of them. We are finding the final temperature in the mixture and that's going to give you 32.5°.. So, if you want to find the heat, for example, you can use that of the water, which is uh 0.499 Specific heat capacity is 4200. So that's a constant, remember CC It's a constant cwc constant. You can just look up, forgot. Uh, so they're changing temperature of water. It's uh, 32.5. That's -25. Okay, so this gives you the heat flow uh, in the system. Yeah, So let's see what that gives by phone on nine tires for 200. Uh, I was 32.5, My was 25. So the heat flow was uh, Somewhere around one 57 I'm stand to the 4th Jews. So this is the heat flow in the system. Thank you very much.


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