Question

how much heat must be added to raise the temperature of 100 gm of water from 5 degree Celsius to 95 degree Celsius. specific heat capacity of water =​

Answer 1

Explanation:

m=100g,C=1calg

−1

o

C

−1

ΔT=(95−5)

o

C=90

o

C

Q=mCΔT

=100g×1calg

−1

o

C

−1

×90

o

C

=100×1×90cal

=9000cal=9kcal