How do I work a specific heat problem?

Answer 1

Specific heat problems are very simple, but you just have to think about them carefully. Here's an example of a specific heat problem that I am just making up for you:

A 10.0 g diamond is heated to 115°C, and is placed in a container containing 100.6 g of liquid water at 25°C. The diamond cools to a final temperature of 30°C. What is the final temperature of the water. In this problem that I have made up, I am telling you that the mass, specific heat, and everything else about the container itself is negligible, so don't worry about it. Also, assume NO heat is lost to the surroundings. The specific heat capacity of water is 4.18 and the specific heat capacity of diamond is 0.509. Don't forget to convert temperature to Kelvin. Let me now explain the concept of the problem:

All that is happening is that heat is being transfered from one place to another. Heat transfered to or from an object is equal to the quantity mcΔT, where m is the mass of the object, c is the specific heat capacity, and ΔT is the change in temperature. ΔT is found using the initial temperature, Ti, and the final temperature, Tf. ΔT can be equal to Tf - Ti OR Ti - Tf. There's an easy way to decide which of the two to use. Use the one that will result in a positive number. For example, you know that the diamond is losing heat. This means that the final temperatue of the diamond will be lower than the initial temperature. If you did Tf - Ti, you would get a negative number. Therefore, you know that you must use Ti - Tf. Sometimes If heat is transfered directly from one object to another without being lost, as in THIS problem, then the mcΔT of one object is equal to the mcΔT of the other object. Now, let's start the problem:

Heat transfered from diamond to water

mcΔT of diamond = mcΔT of water

(10.0)(0.509)(388 - 303) = (100.6)(4.18)(Tf - 298)

432.65 = 420.508Tf - 125311.384

420.508Tf = 125744.034

Tf = 299 K = 26°C

According to the calculations, the final temperature of the water is 26°C (not much change).

I hope this example problem that I made up for you has helped you to understand how to approach and solve specific heat problems. Good luck with future problems.

Answer 2

Keith - it particularly is been a actual long term for me for the reason that chemistry... and that i'm, of course, "winging" this: to improve the temperature of the 250 g of water via 1ºC demands 250 x 4.184 = one million,046 J. To get that plenty warmth out of 155g of Cu the copper will could desire to kick back (one million,046)/(.385 x a hundred and fifty five) = 17.53ºC the ultimate temperature... of direction... would be T. The copper cools 17.fifty 3 circumstances as plenty using fact the water warms the adaptation between the nice and cozy copper and the cool water is, initially, a hundred thirty.2 C tiers. i'm questioning of the temperature hollow remaining in increments of 18.fifty 3 C tiers. that's by way of the fact, for each 1º improve in water temp., there's a 17.53º cut back in the copper's temp... 18.53º closure. a hundred thirty.2/18.fifty 3 = 7.02. there will be 7 of those 1º water temp. will improve; water temp. ends at 26.8º. enable's examine: If the copper cooled a entire of 7 x 17.53º it is going to cool approximately 123.2º. that would have the copper end at a hundred and fifty - 123.2 = 26.8º confident! do not you in basic terms like it whilst it extremely works? Richard

Answer 3

In what form?