6.5 kg chunk of ice at -120C is placed in a 0.22 kg of water at 220C. Ignore any heat flow with the surroundings, including the container, assume the final mixture is ice. Determine the following;
heat lost or gained by water
"c_{ice}m_{ice}(t-t_{ice})=c_{water}m_{water}(t_1-t_2)+\\lambda m_{water}+c_{ice}m_{water}(t-t_{ice})"
"2140\\cdot 6.5\\cdot (t-(-12))=4200\\cdot 0.22\\cdot(22-0)+333500\\cdot 0.22+2140\\cdot 0.22\\cdot(0-t)"
So, we get "t=-5.1\u00b0C" .
"Q=4200\\cdot 0.22\\cdot(22-0)+333500\\cdot 0.22+2140\\cdot 0.22\\cdot(0-(-5.1))="
"=96099.08\\ (J)" . Answer
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