A 271-g sample of nickel at 94.1°C is placed in 100.0 mL of water (density = 1.00 g/mL) at 23.9°C. What is the final temperature of the water?

Question

[Assume that no heat is lost to or gained from the surroundings. Specific heat capacity of nickel = 0.444 J/(g·K) and of water = 4.184 J/(g·K).]

Answer 1

heat lost by Ni + heat gained by H2O = 0
[mass Ni x specific heat Ni x (Tfinal-Tinitial)] + [mass H2O x specific heat H2O x (Tfinal-Tinitial)] = 0

Answer 2

Would I have to rearrange the equation to solve for the final temp of the water?

Answer 3

You can but that's the hard way to do it. It is much easier to substitute the individual numbers and work it out by pieces.

Answer 4

Do I already know the final temp of the nickel because I don't know where to start even though you have given me the equation.

Answer 5

No, of course you don't know the final T. That's what you're trying to find. It ways that in the problem, "What is the final T if ..."

heat lost by Ni + heat gained by H2O = 0
[mass Ni x specific heat Ni x (Tfinal-Tinitial)] + [mass H2O x specific heat H2O x (Tfinal-Tinitial)] = 0
mass Ni = 271 g
specific heat Ni = 0.444
Tinitial Ni = 94.1 C
Tfinal = ?
mass H2O = 100 mL and since water has density of 1.00 g/mL, this is 100 g.
specific heat H2O = 4.184
Tinitial H2O = 23.9 C
Tfinal = ?
Note: Tfinal for Ni and Tfinal for H2O will be the same T since that is where equilibrium will be attained; i.e., when the Ni has lost heat and the H2O has gained heat and both are at the same final temperature.

Answer 6

Okay. Thanks for clarifying the problem for me.