Question

If 24500 J is applied to 125g of water at 35 C, what will the final temperature of the water be?

Answer 1

Answer:

81.9C

Explanation:

δ

H

=

m

×

C

p

×

δ

T

, where  

δ

H

is the enthalpy or change in energy, m is the mass, Cp is the specific heat, and  

δ

T

is the change in temperature.

We know  

δ

H

= 24500J, the specific heat of water is 4.18J/mol*k, and m=125g

So we alter our formula a little bit and substitute in what we know:

δ

T

=

δ

H

m

×

C

p

=

24500

J

125

g

×

4.18

J

m

o

l

×

k

=

46.9

k

Celsius and kelvin have the same scale for units, so simply add the change to the original temperature.

We have: Tfinal = 35+46.9=81.9C

Hope that helps!