A 250-g piece of lead is heated to 100 degree C and is then placed in a 400-g co

Question

A 250-g piece of lead is heated to 100 degree C and is then placed in a 400-g copper container containing 500 g of water. The specific heat of copper is c = 0.386 kJ/kg middot K. The container and water had an initial temperature of 18.0 degree C. When thermal equilibrium is reached, the final temperature of the system is 19.15 degree C. Assuming no heat is lost from the system, what is the specific heat of the lead? (The specific heat of water is 4180 J/kg. K.) a. 0.119 kJ/kg middot K b. 0.128 kJ/kg middot K c. 0.110 kJ/kg middot K d. 0.0866 kJ/kg middot K e. 0.0372 kJ/kg middot K

Explanation / Answer

Given that

mass m1=0.25 kg

mass m2=0.4 kg

mass m3=0.5 kg

specific heat of copper S2=386 J/kg.K

specific heat of water S3=4180 J/kg.K

temperature T1=373 k

temperature T2=18+273=291 k

temperature T3=19.15+273=292.15 k

basing on the concept of calarometer

now we find the specific heat of the lead

m1s1[T1-T3]=[m2s2+m3s3](T3-T2)

0.25*S1*[373-292.15]=[0.4*386+0.5*4180](292.15-291]

20.213*S1=2581.1

specific heat of lead S1=128 J/kg.k=0.128 KJ/kg.k

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