Question

What is the amount of heat, in joules, required to increase the temperature of a 49.5-gram sample of water from 22°C to 66°C?
(1) 9100 J
(2) 4600 J
(3) 1400 J
(4) 2300 J

Answer 1

Answer:

The answer to your question is: (1) 9100 J

Explanation:

Data

Q = ? J

mass = 49.5 g = 0.0495 kg

T1 = 22°C

T2 = 66°C

Cp = 4.18 J/Kg°C

Formula

                   Q = mCpΔT

                   

Substitution

                   Q = (4.95)(4.18)(66 - 22)

                   Q = 9104 J

Answer 2

Answer:

(1) 9100 J.

Explanation:

This would be an increase of 44 degees C and the specific heat of water is 4.186 joules //g/degree C.

So it is 49.5 * 4.186 * 44

= 9100J.