What is the total amount of heat absorbed by 100.0

grams of water when the temperature of the water is

increased from 30.0°C to 45.0°C?

A) 418 J

B) 6270 J

C) 12 500 J

D) 18 800 J

Option (B) 6270J

Explanation:

The following were obtained from the question:

M = 100g

T1 = 30°C

T2 = 45°C

ΔT = 45 - 30 = 15°C

C = 4.18J/g°C

Q=?

Q = MCΔT

Q = 100 x 4.18 x 15

Q = 6270J

Therefore, the total amount of heat absorbed is 6270J