658.9 J of heat was added to 15.05 g of water. The final temperature of the water was 57.8 C. What was the initial temperature of the water?

Answer 1

In this problem, you will use the specific heat of water and the heat transfer equation to solve.

Q=m C ΔT

Q is the heat transferred in J

m is the mass in g

C is the specific heat heat of a substance; liquid water's specific heat is 4.18 J/g ^OC

ΔT is the change in heat in degrees C

You're solving for the initial temperature. Since you know the final temperature, you need to calculate what the change in temperature is.

Q= m C ΔT

658.9 J= 15.05g x 4.18 J/g ^OC x ΔT

658.9 J= 62.91 J/^OC x ΔT

10.5 ^OC= ΔT

This is the change in temperature. Since the initial temperature was lower than the final of 57.8, subtract to find the original:

57.8 ^OC - 10.5^OC = 47.3^OC

47.3^OC was the initial temperature.