Q.

The specific heat of a substance is given by C = a + bT, where a=1.12kJkg1c1&b=0.016kJ kgc1k1. 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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a

235 kJ

b

205 kJ 

c

225 kJ 

d

215 kJ 

answer is C.

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Detailed Solution

Heat required =Q=280312mCdT=280312(1.2)(a+bT)dT=1.2aT+bT22280312

=1.21.12(312280)+0.0162(312)2(280)2=1.21.12×32+0.0162×(9734478400)=1.2[35.84+0.008×18944]=224.8704KJ=225KJ

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