Question

practice heat transfer (6.3 and 6.4)

1. if 150.0 g of zinc at 100.0°c and 250.0 g of liquid water at 10.0°c are mixed in an insulated container, what is the final temperature of the mixture? ( c_{zinc}=390j/kgcdot^{circ}c ) (14.76°c)

Explanation:

Step1: Convert mass units

Convert mass of zinc from grams to kilograms: \(m_{zinc}=150.0\ g = 0.1500\ kg\), mass of water \(m_{water}=250.0\ g=0.2500\ kg\).

Step2: Set up heat - transfer equation

According to the principle of heat transfer \(Q_{lost}=Q_{gained}\). The heat - transfer formula is \(Q = mc\Delta T\). For zinc, \(Q_{zinc}=m_{zinc}c_{zinc}(T_{zinc}-T_f)\); for water, \(Q_{water}=m_{water}c_{water}(T_f - T_{water})\). Here \(c_{water}=4186\ J/kg\cdot^{\circ}C\), \(T_{zinc}=100.0^{\circ}C\), \(T_{water}=10.0^{\circ}C\). So \(m_{zinc}c_{zinc}(T_{zinc}-T_f)=m_{water}c_{water}(T_f - T_{water})\).

Step3: Expand and solve for \(T_f\)

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Substitute the values: \(m_{zinc}=0.1500\ kg\), \(c_{zinc}=390\ J/kg\cdot^{\circ}C\), \(T_{zinc}=100.0^{\circ}C\), \(m_{water}=0.2500\ kg\), \(c_{water}=4186\ J/kg\cdot^{\circ}C\), \(T_{water}=10.0^{\circ}C\).

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Answer:

\(14.76^{\circ}C\)