<span>дано:</span>
<span>С=63Дж/К</span>
<span>m1=250r=0,25kr</span>
<span>T1=12’C=285K</span>
<span>m2=500r=0,5kr</span>
<span>T2=100’C=373K</span>
<span>c2=400Дж</span><span>/</span><span>кг</span><span>К</span>
<span>T</span>=33’<span>C</span>=306<span>K</span>
<span>c</span>1-?
<span>Q</span>1=<span>C</span>(<span>T</span>-<span>T</span>1)- нагрев калориметра
<span>Q</span>2=<span>c</span>1<span>m</span>1(<span>T</span>-<span>T</span>1)-<span>нагрев масла</span> , <span>c</span>1=<span>Q</span>2/<span>m</span>1(<span>T</span>-<span>T</span>1)
<span>Q</span>3=<span>c</span>2<span>m</span>2(<span>T</span>2-<span>T</span>)- остывание медного тела
По закону сохранения Q1+Q2=Q3
<span>Q2=Q3-Q1</span>
<span>Q3=400*0,5*(373-306)=13400</span><span>Дж</span>
<span>Q</span><span>1=63*(306-285)=1323 Дж</span>
<span>Q</span><span>2=13400 Дж-1323 Дж=12077 Дж</span>
<span>C1=12077/0,25*(306-285)=2300Дж/кг К</span>
<span> </span>