Question

It requires the removal of 1.25 kJ of energy to cool a certain sample of pure silver from 15.2 degrees C to 12.0 degrees C. Calculate the mass of the sample of silver?

I don't understand what I'll have to do when REMOVING energy and COOLING a sample. I only understand what to do when I heat it!

Answer 1

Consider that heat is energy.

When the piece of silver cools. Is heat removed or gained from the piece of silver (which we will treat as the system)?

Intuitively heat seems hot to you: when we remove heat, something cools.

Now, recall,

`q = mC_sDeltaT`

where `q` is the heat, and `q<0` in this case due to the preceding reasoning.

Hence,

`m = q/(C_sDeltaT) = (-1.25"kJ" * (10^3"J")/"kJ")/((0.233"J")/("g" * ° "C" ) * -3.2° "C") approx 1.68*10^3"g" * "kg"/(10^3"g") = 1.68"kg"`

of silver are cooled given your data.