Answer :
The final temperature of the coffee and the aluminum cup is approximately 63.49ºC.
To find the final temperature of the coffee and the aluminum cup, we need to use the principle of conservation of energy. Specifically, the heat lost by the coffee will be equal to the heat gained by the aluminum cup. We use the formula for heat transfer:
- [tex]Q = mc\Delta T[/tex]
where [tex]Q[/tex] is the heat transferred, [tex]m[/tex] is the mass, [tex]c[/tex] is the specific heat capacity, and [tex]\Delta T[/tex] is the change in temperature. Let's denote the final temperature by [tex]T_f[/tex].
Heat lost by the coffee:
- [tex]Q_{coffee} = m_{coffee} c_{coffee} (T_{initial, coffee} - T_f)[/tex]
where:
[tex]m_{coffee} = 0.35 \text{ kg}[/tex]
[tex]c_{coffee} = 1300 \text{ J/(kgºC)}[/tex]
[tex]T_{initial, coffee} = 85ºC[/tex]
Heat gained by the aluminum cup:
- [tex]Q_{aluminum} = m_{aluminum} c_{aluminum} (T_f - T_{initial, aluminum})[/tex]
where:
[tex]m_{aluminum} = 0.25 \text{ kg}[/tex]
[tex]c_{aluminum} = 900 \text{ J/(kgºC)}[/tex]
[tex]T_{initial, aluminum} = 20ºC[/tex]
Since the heat lost by the coffee is equal to the heat gained by the aluminum cup:
- [tex]m_{coffee} c_{coffee} (T_{initial, coffee} - T_f) = m_{aluminum} c_{aluminum} (T_f - T_{initial, aluminum})[/tex]
Substituting the values, we get:
- [tex]0.35 \times 1300 \times (85 - T_f) = 0.25 \times 900 \times (T_f - 20)[/tex]
Simplifying the equation:
- [tex]455 \times (85 - T_f) = 225 \times (T_f - 20)[/tex]
Expanding both sides:
- [tex]38675 - 455T_f = 225T_f - 4500[/tex]
Combining terms:
- [tex]38675 + 4500 = 455T_f + 225T_f[/tex]
- [tex]43175 = 680T_f[/tex]
Solving for [tex]T_f[/tex]:
- [tex]T_f = \frac{43175}{680} \approx 63.49^\circ C[/tex]
