Question

The specific heat capacity of a metal is given as 0.745 J/ degree Celcius.What mass of the metal must be heated to 95 degree Celcius in order to be used to warm 2.548 L of water from 18.1 degree Celcius to 22.5 degree Celcius ?


I'm so confused I don't even know where to start
I know that Q= m•s•delta T
I'm thinking to rearrange the equation to
m=q/s•Delta T

I don't know I think what's really throwing me off is the 95degree Celcius

Answer 1

Note the correct spelling of celsius.
You have half of it but not the other half. The metal is heated to 95 C then placed in the water. Then heat lost by the metal + heat gained by the water = 0
heat lost by metal is
q = mass metal x specific heat metal x (Tfinal-Tinitial)

heat gained by the water is
q = mass H2O x specific heat H2O x (Tfinal-Tinitial)

Set the sum of these to zero and you have it.
[mass metal x specific heat metal x Tfinal-Tinitial)] + [mass H2O x specific heat H2O x (Tfinal-Tinitial)] = 0
[mass metal x 0.745 x (22.5-95)] + [2548 x 4.184 x (22.5-18.1)] = 0
Only one unknown.