Дано:
<span>t1</span>=<span>0</span>’<span>C</span>
<span>t</span>2=<span>49,9</span>’<span>C</span>
<span>m1=m</span>
<span>m</span>2=<span>m</span>
<span>с</span>=420<span>0</span>Дж/к<span>гС</span>
<span>L=330000Дж/кг</span>
<span>t-?</span>
<span>Q</span>1=<span>Lm</span>/2- плавление льда массой m/2
<span>Q</span>2<span>=</span><span>c</span><em><span>m</span></em><em><span>∆</span></em><span>t</span><span>-нагрев воды ,</span><em><span><span>∆</span></span></em><span>t</span><span>=</span><span>t</span><span>-</span><span>t</span><span>1</span>
<span>Q</span><span>3=</span><span>cm</span><em><span>∆</span></em><span>t</span><span>2- остывание горячей воды</span><em><span><span> </span></span></em><em><span>∆</span></em><span>t</span><span>2=t2-t</span>
<span>По закону сохранения энергии Q1+Q2=Q3</span>
<span>Lm/2+cm(t-t1)=cm(t2-t)</span>
<span>L/2+c(t-t1)= c(t2-t)</span>
<span>L/2+ct-ct1=ct2-ct</span>
<span>2ct= ct2+ ct1-L/2</span>
<span>t=</span><span>(ct2+ ct1-L/2)/2c</span>
<span>t=[2c(t2+t1)-L]/4c</span>
<span>t=(2*4200*49,9-330000)/4*4200=5’C</span>