Question

2500 cal of heat are absorbed by 150 grams of water that is initially at 23°C, what is the final temperature of the water?

Answer 1

H=MC∆T

Heat = Heat absorbed or given off

M=Mass

C=Specific Heat Capacity of substance (Water in thus case). It has a Specific Heat capacity of 4200J/kg.k

∆T= Temp Change (T2-T1)

I'll solve with My Heat in Joules.

The heat is given in Calorie

1Cal = 4.186J

2500cal = 10,465J.

∆T= H/MC

=10465/0.15(4200)

=16.6°C

∆T=16.6

T2 - T1 = 16.6

T2 = 16.6+T1

T2 = 16.6 + 23

T2 = 39.6

Approx. 40°C