∙A copper pot with a mass of 0.500 kg contains 0.170 kg of water, and both are at a temperature

step 1 Hey everyone, this is question number 44 from Chapter 14. In this problem, we're given a copper

∙A copper pot with a mass of 0.500 kg contains 0.170 kg of...

$\bullet$ A copper pot with a mass of 0.500 $\mathrm{kg}$ contains 0.170 $\mathrm{kg}$ of water, and both are at a temperature of $20.0^{\circ} \mathrm{C} . \mathrm{A} 0.250 \mathrm{kg}$ block of iron at $85.0^{\circ} \mathrm{C}$ is dropped into the pot. Find the final temperature of the system, assuming no heat loss to the surroundings.

Key Concepts

Conservation of Energy

In thermal processes, the conservation of energy principle states that the heat lost by hotter substances is equal to the heat gained by cooler substances when no external heat is exchanged. This principle is crucial when calculating the final equilibrium temperature in a closed system where multiple objects with different initial temperatures interact.

Specific Heat Capacity

Specific heat capacity is a property of a material that indicates the amount of heat needed to raise the temperature of a unit mass of the material by one degree Celsius. It is essential for determining how different substances, such as metals and water, change temperature when absorbing or releasing heat.

Thermal Equilibrium

Thermal equilibrium refers to the state where all parts of a system have reached the same temperature, resulting in no net heat transfer between them. In problems involving objects at different initial temperatures, the final equilibrium temperature is determined by the balance of heat exchange among the components.

Transcript

00:01 Hey everyone, this is question number 44 from chapter 14.

00:04 In this problem, we're given a copper pot of mass 0 .5 kilograms.

00:08 It contains 0 .17 kilograms of water.

00:11 Both of them are at 20 degrees celsius.

00:13 And then an iron block of mass 0 .25 kilograms is dropped into the pot, and it is at 85 degrees celsius.

00:21 So then we're asked to find, that's all of our givens, and we're asked to find the final temperature of the system, and we're told there's no heat loss.

00:28 So again, we're having temperature changes, masses, so we should recognize q equals mc delta t as our go -to equation for this kind of question.

00:41 And we're solving for final temperature, but we need to go ahead and define q for each one of the parts of the system, which would be the copper pot, the water inside the pot, and the iron block.

00:53 So, q for the copper pot, let me draw a better q.

01:02 Qc for copper pot is equal to mass, 0 .5 kilograms.

01:12 C for copper is 390 joules per kilogram k.

01:21 And then for delta t, we know that any change is to find us final minus initial.

01:26 So we're going to see t is our final and minus our initial for the.

01:31 The pot is 20 degrees c.

01:34 Okay, so that's our equation for q of c.

01:36 Now we can go on to q of the water inside the pot is equal to mass of the water, 0 .17 kilograms.

01:56 For water, c is 4190, joules per kilogram k.

02:05 And then our delta t again is going to be, for the water, starts at 20.

02:10 So our final is t.

02:11 We don't know that minus 20 degrees celsius which is our initial and then we can do q of the block which is equal to mass of the block 0 .25 kilograms c for iron is 470 and then our temperature is going to be our final t minus 85 degrees c so the same thing for every single one of them.

03:04 Okay, so we have these three expressions.

03:07 Now we can go ahead and multiply them out and simplify them, and we're told there's no heat transfers...

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