thermal equilibrium

Question

Part A.

What is the energy change when the temperature of 10.4 grams of solid sulfur is decreased from 35.3 °C to 22.4 °C?

Answer:  Joules

Part B.

A chunk of aluminum weighing 18.5 grams and originally at 98.95 °C is dropped into an insulated cup containing 77.6 grams of water at 21.43 °C.

Assuming that all of the heat is transferred to the water, the final temperature of the water is  °C.

Answer 1

Hi, Natalie,

I will use a specific heat for sulfur of 7.15 J/kgK. Then use the equation q = cMDt, where q is the energy in Joules, c is the specific heat in J/kgK, M is the mass in kg, and Dt is the change in degrees Kelvin. Since the specific heat has units of kg, we must convert the 10.4 grams of sulfur to 0.0104 kg of sulfur. The Dt, or change in temperature is -12.9 C, which is also -12.9K.

Then use the formula with these numbers to get the Joules released by the sulfur when it cools down:

q (J) = (7.15 J/kgK)*(0.0104kg)*(-12.9K)

This yields -0.959 J. Negative, because the heat left the sulfur.

The next problem is handled much the same way. I used a specific heat for Al of 0.9 J/gK. Since this unit is in grams, we can use the 18.5 grams mass provided.

The energy given up by the Al is q (J) = (0.9J/gK)*(18.5 g)(-77.52K), This leads to -1,291 J that is released by the aluminum. To find the temperature increase of the water, we use the same equation but with the c of water equal to 4.18 J/gC. The enegy going into the water is the same as the energy coming from the aluminum, or 1,291J.

1,291J = 4.186J/gC *(77.6g)*(Dt), where Dt is the change in temperature from 21.43C. I get a temperature increase of 3.97C, brining the water temperature up to 25.40C.

I hope this helps,

Bob