Question

A metal smith pours 2.4kg of lead shot at 91.9 degrees C into 1.0kg kg of water at 21.4 degrees C in an isolated container. What is the final temperature of the mixture? Give your answer to one decimal place in units of K.

C of water = 4.19kJ/(kg*K), C of lead = 0.129kJ/(kg*K)

Answer 1

Here lead shot is at higher temperature and water is at lower temperature.

when lead is poured into water heat is lost. The lost heat is gained by water.Since it is an isolated system no energy is lost.

$q_{lost}=q_{gained}$

$m_{lead}*C_{lead}*\Delta T_{lead}=m_{water}*C_{water}*\Delta T_{water}$

$2.4kg*0.129kJ/kgK*(T_{lead}-T_{final})=1kg*4.19kJ/kgK*(T_{final}-T_{water})$

$2.4*0.129*(365.05K-T_{final})=1*4.19*(T_{final}-294.55K)$

$(365.05K-T_{final})=1*4.19*(T_{final}-294.55K)/(2.4*0.129)$

$(365.05K-T_{final})=13.534*(T_{final}-294.55K)$

$365.05-T_{final}=13.534*T_{final}-3986.4397$

$4351.4897=14.534*T_{final}$

ANSWER; ${\color{Red} T_{final}=299.4K}$

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