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Q.

An ice cube of mass 0.1 kg at 00C is placed in an isolated container which is at 2270C. The specific heat S of the container varies with temperature T according to the empirical relation S = A+BT, where A = 100 cal/kg-K and B = 2×10-2cal/kg-K2. If the final temperature of the container is 270C,  find the mass of the container.

(Latent heat of fusion for water = 8×104 cal/kg, specific heat of water = 103 cal/kg-K)

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a

0.621 kg

b

0.495 kg

c

0.224kg

d

0.336 kg

answer is A.

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Detailed Solution

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Heat received by ice is

    Q1 = mL+mCT = 10700 cal. 

Heat lost be the container is

Q2 = 300500mC(A+BT)dT  = mC[AT+BT22]300500

                                                = 21600 mC

By principle of calorimetry, Q1 = Q2

 mc = 0.495 kg

 

 

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