Problem

Two substances, A and B, initially at different t…

01:56

Problem 62 Hard Difficulty

We submerge a 2.85-g lead weight, initially at 10.3 C, in 7.55 g of water at 52.3 C in an insulated container. What is the final temperature of both substances at thermal equilibrium?

Answer

$$
T_{\mathrm{f}}=51.83^{\circ} \mathrm{C}
$$

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Video Transcript

So in this problem we have two point eight five grams of lead at ten point three degrees Celsius, being submerged in seven point five five grams ofwater at fifty two point three degrees Celsius. And we want to find the final temperature off this system at equilibrium. And so the way we do that is we can use the formula. Q. Gained equals Q lost. And so the amount of heat gained by the lead will be the same as the amount of heat lost by the water. And so we can ahh expand these formulas out a little bit. So we have the mass of one, which is the lead, the specific heat of the lead, and then the delta t of the lead equals mass to see two delta T too. So the mass of the water, the specific heat and the change in temperature of the water. And so now we can plug in our values. We have two point eight five grams of lead and the specific heat of lead is zero point one two eight and then our delta t we're going to take our tea final what we're looking for minus the ten point three degrees Celsius that the lead starts out. Now we can plug in our massive water seven point five five grams times the specific heat of water four point one eight. And now we want to take our fifty two point three degrees Celsius and subtract our final temperature. And the reason we do it in this way is that the final temperature will be somewhere between ten point three and fifty two point three. And we want both of these Delta teas to be positive. And so in the first one, we have to subtract. And in the second one we have tio subtract the final temperature instead of the known temperature. And so now that we have this, we can ah rearrange these to solve for t f. And so the first step of this we get zero point three six for eight times t f minus three point seven five seven equals sixteen, fifty point five three six minus thirty one point five five nine times t f. And now we can combine like terms. I'll go to a new page for this So we end up with thirty one point nine two three eight t F equals sixteen fifty four point two nine three. And when we divide the thirty one point nine two three eight we get, our final temperature is fifty one point eight degrees Celsius, and that is our final answer.

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