Question

A sample of marshmallow is combusted directly below a can containing 100g, of water initially at25.0°C. The reaction releases 2.1 kJ of heat energy, all of which is transferred to the water. What is the final temperature of the water? (The specific heat of water is 4.2 J/ (g"C)

Answer 1

$q=mc$ΔT

Where,

q= Heat Energy =2.1kJ

m= Mass =100g

c= Specific Heat = 4.2 J/ (g°C)

ΔT= Temperature change

Changing kilojoules (kJ) To Joules(J),

1kJ=1000J

2.1kJ×1000=2100J

Let initial temperature be T_i and final temperature be T_f

T_i =25°C

Therefore from the formula $q=mc$ΔT;

ΔT= mc/q

ΔT=100g×4.2 J/(g°C)/2100J

ΔT=0.2°C

ΔT=T_f − T_i

T_i =25°C

0.2°C=T_f − 25°C

T_f =25.2°C

The final temperature of the water is 25.2°C.