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.

Correct answer:

t₂ =  28.3452 °C

Step-by-step explanation:

$$\begin{gathered} m_1=0.4\text{kg} \\ {t}_1=250\text{°C} \\ {V}_2=0.4\text{l} \\ t=35\text{°C} \\ {c}_1=130\text{J/kg/°C} \\ {c}_2=4200\text{J/kg/°C} \\ {\rho }_2=1\text{kg/l} \\ {m}_2={\rho }_2\cdot V_2=1\cdot 0.4=\frac{2}{5}=0.4\text{kg} \\ {Q}_1=Q_2 \\ {m}_1\cdot c_1\cdot (t_1-t)=m_2\cdot c_2\cdot (t-t_2) \\ 0.4\cdot 130\cdot (250-35)=0.4\cdot 4200\cdot (35-t_2) \\ 1680t_2=47620 \\ {t}_2=\frac{47620}{1680}=28.3452381 \\ {t}_2=\frac{2381}{84}\doteq 28.345238=28.3452\text{°C} \end{gathered}$$

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