An aluminum object with a mass of 2.45 kg at a temperature of 26.1 °C comes into thermal contact with a 9.89 kg copper object initially at a temperature of 98.2 °C. What will be the equilibrium temperature of the two objects? Neglect heat transfer between the objects and the environment.

The specific heats are:
- [tex]c_{\text{Al}} = 900 \, \text{J/kg°C}[/tex]
- [tex]c_{\text{Cu}} = 387 \, \text{J/kg°C}[/tex]

(Answer in °C)

To find the equilibrium temperature, we need to use the principle of conservation of energy. The heat gained by the aluminum object is equal to the heat lost by the copper object.

The formula for heat transfer is Q = m c Δ T, where m is mass, c is specific heat, and ΔT is change in temperature.

Let T-f be the final temperature of both objects.

For the aluminum object:

Q = m c Δ T

Q = (2.45 kg)(900 J/kg°C)(T-f - 26.1 °C)

For the copper object:

Q = mcΔT

Q = (9.89 kg)(387 J/kg°C)(26.1 °C - T-f)

Setting the two quantities equal to each other and solving for T-f:

(2.45 kg)(900 J/kg°C)(T-f - 26.1 °C) = (9.89 kg)(387 J/kg°C)(26.1 °C - T-f)

T-f = (9.89 kg)(387 J/kg°C)(26.1 °C) / [(2.45 kg)(900 J/kg°C) + (9.89 kg)(387 J/kg°C)] + 26.1 °C

T-f = 45.6 °C

Therefore, the equilibrium temperature of the two objects is 45.6 °C.