Question

Heat Transfer, Specific Heat. and CalorimetryA 1.28-kg sample of water at 10.0 "C is in a calorimeter. You drop a piece of steel with a mass of 0.385 kg at 215 "C into it.After the sizzling subsides, what is the final equilibrium temperature? (Make the reasonable assumptions that any steamproduced condenses into liquid water during the process of equilibration and that the evaporation and condensation don’taffect the outcome. as we’ll see in the next section.)

Answer 1

Answer:

$T_f=16.975\ ^{\circ}C$

Explanation:

Given:

• mass of water, $m_w=1.28\ kg$

• initial temperature of water, $T_{iw}=10\ ^{\circ}C$

• mass of steel, $m_s=0.385\ kg$

• initial temperature of steel, $T_{is}=215\ ^{\circ}C$

• specific heat capacity of steel, $c_s=490\ J.kg^{-1}.K^{-1}$

• specific heat capacity of water, $c_w=4184\ J.kg^{-1}.K^{-1}$

Now by the law of conservation of energy, the heat released by the steel is equal to the heat absorbed by the water:

$Q_s=Q_w$

$m_s.c_s.(T_{is}-T_f)=m_w.c_w.(T_f-T_{iw})$

$0.385\times 490\times (215-T_f)=1.28\times 4184\times (T_f-10)$

$40559.75-188.65\times T_f=5355.52\times T_f-53555.2$

$T_f=16.975\ ^{\circ}C$