Question

Brewed coffee is often too hot to drink right away. You can cool it with an ice cube, but this dilutes it. Or you can buy a device that will cool your coffee without dilution - a 230 g aluminum cylinder that you take from your freezer and place in a mug of hot coffee.

Answer 1

Answer:

$75.5^oC$

Explanation:

Full question:

a. If the cylinder is cooled to $-20^oC$, a typical freezer temperature, and then dropped into a large cup of coffee (essentially water, with a mass of 500 g) at $85^oC$, what is the final temperature of the coffee?

We have a classical heat exchange problem. According to the second law of thermodynamics, heat flows from hotter objects to colder ones, and this means that the coffee cup will lose heat and the cylinder will gain it. We treat the coffee as water. It will lose heat given by the equation:

$Q_1 = c_cm_c(t_{i,c}-t_f)$

Here:

$c_c$ is the specific heat capacity of water, $4.184 \frac{J}{g^oC}$;

$m_c$ is the mass of the coffee, 500 g;

$t_{i,c}$ is the initial temperature of the coffee, $85^oC$;

$t_f$ is the final temperature of the mixture.

Now, aluminum will gain heat given by the equation:

$Q_2 = c_{Al}m_{Al}(t_f-t_{i,Al})$

Here:

$c_{Al}$ is the specific heat of aluminum, $0.900 \frac{J}{g^oC}$;

$m_{Al}$ is the mass of the cylinder, 230 g;

$t_{i,Al}$ is the initial temperature of the cylinder, $-20^oC$;

$t_f$ is the final temperature of the mixture.

Since the energy is conserved, heat lost by the coffee should be equal to the heat gained by the cylinder:

$Q_1 = Q_2$

Substituting the two equations:

$c_cm_c(t_{i,c}-t_f) = c_{Al}m_{Al}(t_f-t_{i,Al})[\tex]

Solve for the final temperature:

[tex]c_cm_ct_{i,c} - c_cm_ct_f = c_{Al}m_{Al}t_f - c_{Al}m_{Al}t_{i,Al}$

$t_f(c_{Al}m_{Al} + c_cm_c) = c_cm_ct_{i,c} + c_{Al}m_{Al}t_{i,Al}$

$\therefore t_f = \frac{c_cm_ct_{i,c} + c_{Al}m_{Al}t_{i,Al}}{c_{Al}m_{Al} + c_cm_c}$

Substitute the numbers:

$t_f = \frac{4.184 \frac{J}{g^oC}\cdot 500 g\cdot 85^oC + 0.900 \frac{J}{g^oC}\cdot 230 g\cdot (-20^oC)}{0.900 \frac{J}{g^oC}\cdot 230 g + 4.184 \frac{J}{g^oC}\cdot 500 g} = 75.5^oC$