We often use a spoon to cool coffee by stirring. Some of the heat is conducted from the coffee to the air by this method, but the spoon itself will cool the coffee a little. If a 12 gram aluminum teaspoon at 22 o C is placed in 175 ml of water (coffee's specific heat is close to water's) at 85 o C, what will the final temperature be? (Other factors, such as heat transfer to the air, may be ignored.)

Question

We often use a spoon to cool coffee by stirring. Some of the heat is conducted from the coffee to the air by this method, but the spoon itself will cool the coffee a little. If a 12 gram aluminum teaspoon at 22 o C is placed in 175 ml of water (coffee's specific heat is close to water's) at 85 o C, what will the final temperature be? (Other factors, such as heat transfer to the air, may be ignored.)

Answer 1

To determine the final temperature of the water and teaspoon, we need to use the principle of conservation of energy.

The equation for heat transfer is given by:

Q = mcΔT

Where Q is the heat transferred, m is the mass of the object, c is the specific heat capacity, and ΔT is the change in temperature.

First, we can calculate the heat transferred from the water to the teaspoon.

For the water:
Q_water = m_water * c_water * ΔT_water

Where
m_water = mass of water = 175 g
c_water = specific heat capacity of water = 4.18 J/g°C (approximately)
ΔT_water = change in temperature of water = final temperature - initial temperature = final temperature - 85°C

For the teaspoon:
Q_teaspoon = m_teaspoon * c_teaspoon * ΔT_teaspoon

Where
m_teaspoon = mass of teaspoon = 12 g
c_teaspoon = specific heat capacity of aluminum = 0.897 J/g°C (approximately)
ΔT_teaspoon = change in temperature of teaspoon = final temperature - initial temperature = final temperature - 22°C

According to the principle of conservation of energy, the heat transferred from the water to the teaspoon will be equal to the heat gained by the teaspoon.

Q_water = Q_teaspoon

m_water * c_water * ΔT_water = m_teaspoon * c_teaspoon * ΔT_teaspoon

Substituting the given values, we have:

175 g * 4.18 J/g°C * (final temperature - 85°C) = 12 g * 0.897 J/g°C * (final temperature - 22°C)

Simplifying the equation:

732.1 (final temperature - 85) = 10.764 (final temperature - 22)

Now we can solve for the final temperature.

732.1 final temperature - 62349.35 = 10.764 final temperature - 237.408

721.336 final temperature = 62112.942

final temperature ≈ 86.1°C

Therefore, the final temperature of the water and teaspoon will be approximately 86.1°C.