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

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235 kJ

205 kJ

225 kJ

215 kJ

answer is C.

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Detailed Solution

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

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

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