Question

A hot iron ball is dropped into a 200. g sample of water initially at 50°C. If 8.4 kJ of heat is transferred from the ball to the water, what is the final temperature of the water? (The specific heat of water is 4.2 J/(g·°C).

Answer 1

Answer: The final temperature of water is 60°C

Explanation:

To calculate the final temperature of water, we use the equation:

$q=mc\Delta T$

where,

q = amount of heat absorbed = 8.4 kJ = 8400 J  (Conversion factor:  1 kJ = 1000 J)

m = mass of water = 200 g

c = specific heat capacity of water = 4.2 J/g.°C

$\Delta T$ = change in temperature = $T_2-T_1=T_2-50^oC$

Putting values in above equation, we get:

$8400J=200g\times 4.2J/g.^oC\times (T_2-50^oC)\\\\T_2=60^oC$

Hence, the final temperature of water is 60°C