250.0 g of water, which has a specific heat capacity of 4.18 J/g°C, is given 4,500 J of heat. If the initial temperature of the water is 25.0°C, what is the fin

Question

I'm not exactly sure how I should approach this. 

Answer 1

The equation for specific heat is

Q = mcΔT

 

You're given Q = 4500J, m = 250g c=4.18 J/gºC, and T_i=25ºC.

Remember that ΔT = T_f - T_i

 

Substitute all of this into the equation, and you should be able to solve for T_f.

Q = mc(Tf - Ti)