An ice cube of mass 0.1 kg at 0°C is placed in an isolated container which is at 227°C. The specific heat s of the container varies with temperature T according to the empirical relation s = A + BT, where A = 100 cal kg-1 K-1 and B = 2 × 10-2 cal kg-1. If the final temperature of the container is 27°C, the mass of the container is (Latent heat of fusion of water = 8 × 104 cal kg-1, specific heat of water = 103 cal kg-1 K-1)

Question

An ice cube of mass 0.1 kg at 0°C is placed in an isolated container which is at 227°C. The specific heat s of the container varies with temperature T according to the empirical relation s = A + BT, where A = 100 cal kg-1 K-1 and B = 2 × 10-2 cal kg-1. If the final temperature of the container is 27°C, the mass of the container is (Latent heat of fusion of water = 8 × 104 cal kg-1, specific heat of water = 103 cal kg-1 K-1) (A) 0.495 kg (B) 0.595 kg (C) 0.695 kg (D) 0.795 kg

Tardigrade Admin · Accepted Answer

Heat lost by container =-∫ limits300500 mC(A+BT)dT =-mC[AT+(BT2/2)]300500=21600 mC Heat gained by ice =miceL+miceswater Δ T =0.1 ×8 × 104+0.1 ×103× 27 =10700 cal According to principle of calorimetry Heat lost by container = Heat gained by ice 21600mC = 10700 or mC=0.495 kg

$0.495 k g$

$0.595 k g$

$0.695 k g$

$0.795 k g$

0.495kg

0.595kg

0.695kg

0.795kg

Heat lost by container =−500∫300​mC​(A+BT)dT =−mC​[AT+2BT2​]500300​=21600mC​ Heat gained by ice =mice​L+mice​swater​ΔT =0.1×8×104+0.1×103×27 =10700cal According to principle of calorimetry Heat lost by container = Heat gained by ice 21600mC​=10700 or mC​=0.495kg

$0.495 k g$

$0.595 k g$

$0.695 k g$

$0.795 k g$