## 4a.9

$\Delta U=q+w$

Hannah_1G
Posts: 100
Joined: Wed Sep 18, 2019 12:22 am

### 4a.9

What formula is used to calculate the change in temperature from copper at 100 degrees C placed in water at 22 degrees C? I don't know how to begin this problem

AKhanna_3H
Posts: 104
Joined: Wed Sep 18, 2019 12:19 am

### Re: 4a.9

This question would use the heat capacity formula q = nCdeltaT. You would solve for q of copper and q of water by setting them equal to each other and assigning a negative sign to the q of water (calorimeter). You can set them equal to each other because the heat lost by one reaction is gained by the other.

Pegah Nasseri 1K
Posts: 100
Joined: Wed Feb 27, 2019 12:15 am

### Re: 4a.9

You would use the equation q = m*c*delta T where q is the heat energy, m is the mass of the copper, c is the specific heat of the copper and delta T is the change in temperature. You know that since no energy is lost to the surroundings, the heat lost by copper will equal the negative value of the heat gained by water. Thus you can set the two equations, the one modeling water and the one modeling copper, to each other:

m_copper * c_copper * delta T = - (m_water * c_water * delta T).

The final temperature of the water and the copper will be the same so you will need to plug in the values that are given to you and solve for the final temperature.

BritneyP- 2c
Posts: 101
Joined: Sat Sep 14, 2019 12:15 am

### Re: 4a.9

you would use q=ncdeltaT, and set copper to equal the negative value for the q for water. Whatever heat is lost be one system is absorbed into the other system.