A pan containing 40 grams of water was allowed to cool from a temperature of 91.0C. If the amount of heat repressed is 1,300 joules what is the approximate final temperature of the water.

74 C

78C

81 C

83C

83°C

Explanation:

The following were obtained from the question:

M = 40g

C = 4.2J/g°C

T1 = 91°C

T2 = ?

Q = 1300J

Q = MCΔT

ΔT = Q/CM

ΔT = 1300 / (4.2x40)

ΔT = 8°C

But ΔT = T1 - T2 (since the reaction involves cooling)

ΔT = T1 - T2

8 = 91 - T2

Collect like terms

8 - 91 = - T2

- 83 = - T2

Multiply through by - 1

T2 = 83°C

The final temperature is 83°C