Question 4.16: Problem: A 1 cm diameter copper sphere heated to 300 °C is d...
Problem: A 1 cm diameter copper sphere heated to 300 °C is dropped into a well‐insulated beaker containing 10 ml of water at 25 °C. What is the final temperature once the water and metal have come to equilibrium?
Find: Final temperature T_2 .
Known: Initial temperature of copper T_{1,Cu} = 300 °C, initial temperature of water T_{1,W} = 25 °C, diameter of copper sphere D = 0.01 m, volume of water V_W = 10 \ ml =10^{–5} \ m^3 .
Diagram:
Assumptions: Water and copper are both incompressible, there is no heat loss from the beaker so Q_{12} = 0.
Governing Equation:
First Law Q_{12}+W_{12}=\Delta U_{12}
Properties: Copper density and specific heat ρ_{Cu} = 8900 kg / m³ and c_{Cu} = 0.386 kJ / kg°C (Appendix 2), water density and specific heat ρ_{W} = 997 kg / m³ and c_{W} = 4.18 kJ / kg°C (Appendix 3).
Mass of copper sphere is
m_{Cu}=\rho _{Cu}\frac{\pi D^3}{6} =8900 \ kg/m^3 \times \frac{\pi (0.01 \ m)^3}{6} =4.6600 \times 10^{-3} \ kg.
Mass of water is
m_W=\rho _WV_W=997 \ kg/m^3 \times 10^{-5} \ m^3=9.97 \times 10^{-3} \ kg.
Since the system is incompressible, W_{12} = 0 so,
\underbrace{Q_{12}}_{=0} +\underbrace{W_{12}}_{=0} =\Delta U_{12}=0
\Delta U_{12}=m_{Cu}\Delta u_{12,Cu}+m_{W}\Delta u_{12,W}=m_{Cu}c_{Cu}(T_2-T_{1,Cu})+m_{W}c_{W}(T_2-T_{1,W})=0.
Substituting into the energy balance,
\Delta U_{12}= 4.66 \times 10^{-3} \ kg \times 0.386 \ kJ/kg^\circ C \times (T_2-300 \ ^\circ C) + 9.97 \times 10^{-3} \ kg \times 4.18 \ kJ/kg^\circ C \times (T_2-25 \ ^\circ C)=0.
Solving,
T_2 =36.379 \ º C.
Answer: The final temperature of the system is 36.4 °C.