Question

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    

Answer 1

$c_{ice}m_{ice}(t-t_{ice})=c_{water}m_{water}(t_1-t_2)+\lambda m_{water}+c_{ice}m_{water}(t-t_{ice})$

$2140\cdot 6.5\cdot (t-(-12))=4200\cdot 0.22\cdot(22-0)+333500\cdot 0.22+2140\cdot 0.22\cdot(0-t)$

So, we get $t=-5.1°C$ .

$Q=4200\cdot 0.22\cdot(22-0)+333500\cdot 0.22+2140\cdot 0.22\cdot(0-(-5.1))=$

$=96099.08\ (J)$ . Answer