1) If a rod of chromium weighing about 323.98 grams is cooled in a bucket containing 35 grams of water, and the final temperature is.

54.5°C, what is the specific heat of chromium?

See explanation below

Explanation:

In order to do this, you are missing two important data, which are the initial temperatures of the water and the rod.

For practical purposes, I will assume this data, and then you'll just replace the procedure with your data to get an accurate result.

So, for this exercise, I will assume that the rod of Chromiun was heat to 100 °C and then cooled with water at 20 °C, so the final temperature is 54.5 °C.

To get the specific heat, we need to use the following expression:

q = m.C.ΔT  (1)

Where:

q: Heat of the substance/compound

m: mass

C: specific heat

ΔT: Difference of temperature of the mixture.

With this expression, we need to calculate first the heat of the water, because:

q₁  + q₂ = 0

Therefore, with the value of the transfered heat, we can determine later the specific heat. So, As we have the specific heat of water which is 4.18 J/g °C, and the temperature and mass, we can determine the heat of the mixture:

ΔT = 54.5 - 20 = 34.5 °C

q₁ = 35 * 4.18 * 34.5

q₁ = 5047.35 J

Now, we can determine the specific heat of the rod:

q₁ = -q₂

-5047.35 = 323.98 * (54.5 - 100) * C

-5047.35 = -14741.09C

Hope this helps