Ahot iron ball is dropped into a 200 g sample of water initially at 50 degrees celsius.

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/gc).a. 40 c
b. 50 c
c. 60 c
d. 70 c

The final temperature is 60 °C

Explanation:

Step 1: Data given

Mass of the iron ball = 200 grams

Initial temperature = 50 °C

8.4 kJ of heat is transferred from the ball to the water.

The specific heat of water = 4.2 J/g°C

Step 2: Calculate the final temperature

Q = m*c*ΔT

⇒ with Q = heat is transferred from the ball to the water = 8400 J

⇒ m = the mass of the ball = 200 grams

⇒ c = the specific heat of water = 4.2 J/g°C

⇒ ΔT = T2 - T1 = T2 - 50

8400 J = 200g*4.2 J/g°C * (T2-50°C)

8400  = 840T2 - 42000

50400 = 840T2

T2 = 50800/840

T2 = 60 °C

The final temperature is 60 °C

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