AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

The specific heat of a substance is given by C = a + bT, where $a = 1.12 kJ {kg}^{- 1} c^{- 1} & b = 0.016 kJ - {kgc}^{- 1} k^{- 1}$ . The amount of heat required to raise the temperature of 1.2 kg of the material from 280 K to 312 K is

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

235 kJ

205 kJ

225 kJ

215 kJ

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Heat required $= Q = \int_{280}^{312} mCdT = \int_{280}^{312} (1.2) (a + bT) dT = 1.2 {[aT + \frac{{bT}^{2}}{2}]}_{280}^{312}$

$\begin{array}{l} = 1.2 [1.12 (312 - 280) + \frac{0.016}{2} \{(312)^{2} - (280)^{2}\}] \\ = 1.2 [1.12 \times 32 + \frac{0.016}{2} \times (97344 - 78400)] = 1.2 [35.84 + 0.008 \times 18944] = 224.8704 KJ = 225 KJ \end{array}$

Watch 3-min video & get full concept clarity

courses

No courses found

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Get Expert Academic Guidance – Connect with a Counselor Today!