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 the water. The mass heat capacity of lead is 130 J/kg/°C. The mass heat capacity of water is 4200 J/kg/°C.

Final Answer:

t2 =  28.3452 °C

Step-by-step explanation:

m1=0.4 kg t1=250 °C V2=0.4 l t=35 °C c1=130 J/kg/°C c2=4200 J/kg/°C ρ2=1 kg/l  m2=ρ2 V2=1 0.4=52=0.4 kg  Q1=Q2  m1 c1 (t1t)=m2 c2 (tt2) 0.4 130 (25035)=0.4 4200 (35t2)  1680t2=47620  t2=168047620=28.3452381  t2=84238128.345238=28.3452°C



Help us improve the example. If you find a mistake, let us know. Thank you!







Tips for related online calculators
Do you know the volume and unit volume, and want to convert volume units?
Tip: Our Density units converter will help you convert density units.
Do you want to convert mass units?

You need to know the following knowledge to solve this word math problem:

Related math problems and questions: