(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

In a physics lab experiment a student immersed 191 one-cent coins (each having a mass of 3.00 g) in boiling water, at a temperature of 100 degrees C. After they reached thermal equilibrium, she fished them out and dropped them into an amount of water of mass 0.275 kg at a temperature of 17.0 degrees C in an insulated container of negligible mass.

What was the final temperature of the coins? (One-cent coins are made of a metal alloy - mostly zinc - with a specific heat capacity of 390 J/(kg * K).)

Use 4190 J/(kg* K) for the heat capacity of the water.

2. Relevant equations

Q=mc(deltaT)

Q1 + Q2=0

3. The attempt at a solution

I'm not sure where you can get a workable system of equations from this problem. I tried finding the Q of the coins, which I found to be Q=(191*.003kg)(390)(T-100) where T=the final temperature of the coins. And the Q of Water: Q=(.275)(4190)(T-17).

You also know that the Q of water and the Q of the coins must be equal to 0 because they are in an insulated container.

Because the water is heated and the coins are cooled, you know that the Q for water is positive and that Qwater=Qcoins, However, they have different final temperatures and I'm not sure how to find one in terms of the other so I can solve the equation I've come up with. Please help!

**Physics Forums - The Fusion of Science and Community**

# Thermodynamics: Q=mc(deltaT)

Know someone interested in this topic? Share a link to this question via email,
Google+,
Twitter, or
Facebook

- Similar discussions for: Thermodynamics: Q=mc(deltaT)

Loading...

**Physics Forums - The Fusion of Science and Community**