Question

The specific heat of a substance is given by C = a + bT, where a = 1.12 kJ kg^-1K^-1 and b = 0.016 kJ-kg K^-2. The amount of heat required to raise the temperature of 1.2 kg of the material from 280 K to 312 K is ______.

Options

• 205 kJ

• 215 kJ

• 225 kJ

• 235 kJ

Answer 1

The specific heat of a substance is given by C = a + bT, where a = 1.12 kJ kg^-1K^-1 and b = 0.016 kJ-kg K^-2. The amount of heat required to raise the temperature of 1.2 kg of the material from 280 K to 312 K is 225 kJ.

Explanation:

Heat required

Q = `int_280^312 mCdT = int_280^312 1.2(a + bT)dt`

= `1.2[aT + (bT^2)/2]_280^312`

= `1.2[1.12(312 - 280) + 0.016/2{(312)^2 - (280)^2}]`

= `1.2[1.12 xx 32 + 0.016/2 xx (97344 - 78400)]`

= 1.2 [35.84 + 0.008 × 18944]

= 224.8704 KJ

= 225 KJ