Heat transfer

We placed a lead object weighing 0.4 kg and 250°C in 0.4 L water. What was the initial temperature of water t2 if the object's temperature and the water after reaching equilibrium was 35°C? We assume that the heat exchange occurred only between the lead object and water. The mass heat capacity of lead is 130 J/kg / °C. The mass heat capacity of water is 4200 J/kg / °C.

Correct answer:

t2 =  28.3452 °C

Step-by-step explanation:

m1=0.4 kg t1=250V2=0.4 l t=35c1=130 J/kg/c2=4200 J/kg/ρ2=1 kg/l  m2=ρ2 V2=1 0.4=25=0.4 kg  Q1=Q2   m1 c1 (t1t)=m2 c2 (tt2) 0.4 130 (25035)=0.4 4200 (35t2)  1680t2=47620  t2=23818428.345238=28.3452C



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