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Solution we have put 214 kg of solid laurasium into a 3.8 kg of liquid te thium at gi°c. heat lost by liquid te thium heat gain by laurasium mq AT = mecera Tot me to the ce ate mt = 3.8 where, where, 4 7 2330 I likg. oc 3.8 kg and AT [97°– TA] Tf » final equilibrium temp. and me • 4 kg, Ces 1165 J 1kg.
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Cc. and AT 22.6 - 6a 16.6. and Le 25149 Ilkg x che = 3495 Tlk.oC ste [TB - 22°C] = 2.4 Norco put all these values in equation ④ 3.8 x 2 330 (97-T;) = 2.4*116 5 x 16 + 2.4 X 25,149 + 2.4 x 349 5x [ T1 - 22°c.] 85%,838 – 8,854 TF 44736_t 60,357.6. +8,388 TH 184,536 938,280.4 17 242 Tf ) It 938, 280.4 17 242 IM TE 54.4°C Ave.
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