A restaurant serves tea in aluminum mugs. The aluminum mug has a mass of 125 g, initially at 24.0 oC with 341 mL tea at 80.0oC. What is the final temperature of the tea after the tea and mug attain equilibrium. Assume the tea has the same specific heat capacity and density as water and that there is no heat exchange with the surroundings.
heat of fusion = 3.95 kJ/mol C(solid) = 0.902 J/goC
heat of vaporization = 10.52 kJ/mol C(liquid) = 0.860 J/goC
boiling point =2470 oC C(gas) = 1.05 J/goC
melting point = 660.oC
To find the final temperature of the tea after it attains equilibrium with the aluminum mug, we need to calculate the heat transfer between the tea and the mug.
First, calculate the initial heat of the tea using the specific heat capacity formula:
Q_initial = m_initial * c * ΔT_initial
where:
m_initial = mass of tea = volume of tea * density of water
ΔT_initial = 80.0 oC - 24.0 oC
c = specific heat capacity of water = 4.18 J/g oC
m_initial = 341 mL * 1 g/mL = 341 g
ΔT_initial = 80.0 oC - 24.0 oC = 56.0 oC
Q_initial = 341 g * 4.18 J/g oC * 56.0 oC = 79973.12 J
Next, calculate the heat transferred to the aluminum mug during the cooling process. We can assume that the heat transferred from the tea to the mug is equal to the heat transferred from the mug to the tea.
Q_mug = Q_initial
Now, calculate the heat released by the mug using the formula:
Q_mug = m_mug * c_mug * ΔT_mug
where:
m_mug = mass of the aluminum mug = 125 g
ΔT_mug = final temperature - initial temperature (unknown)
c_mug = specific heat capacity of aluminum = 0.902 J/g oC
Q_mug = 125 g * 0.902 J/g oC * ΔT_mug
Since Q_mug = Q_initial, we can set the two equations equal to each other:
125 g * 0.902 J/g oC * ΔT_mug = 79973.12 J
Solving for ΔT_mug:
ΔT_mug = 79973.12 J / (125 g * 0.902 J/g oC)
ΔT_mug = 71.0 oC
Now, calculate the final temperature of the tea:
Final temperature = initial temperature + ΔT_mug
Final temperature = 24.0 oC + 71.0 oC
Final temperature = 95.0 oC
Therefore, the final temperature of the tea after attaining equilibrium with the aluminum mug is 95.0 oC.
To find the final temperature of the tea after it attains equilibrium with the aluminum mug, we need to consider the heat exchange between the tea and the mug.
The heat gained or lost by a substance can be calculated using the formula:
Q = mcΔT
where Q is the heat gained or lost, m is the mass of the substance, c is the specific heat capacity of the substance, and ΔT is the change in temperature.
In this case, we have two substances - the aluminum mug and the tea.
Let's calculate the heat gained or lost by the aluminum mug first:
Q_mug = mcΔT_mug
m_mug = 125 g (given mass of aluminum mug)
c_mug = c_water (specific heat capacity of water = 4.18 J/goC)
ΔT_mug = T_final - T_initial (change in temperature of the mug)
Now, the heat gained or lost by the tea is:
Q_tea = mcΔT_tea
m_tea = 341 mL (given volume of tea)
Since tea has the same density as water, we can convert the volume to mass assuming the density of water is 1 g/mL.
m_tea = 341 g
c_tea = c_water (specific heat capacity of water)
ΔT_tea = T_final - T_initial (change in temperature of the tea)
Since the tea and the mug are in thermal equilibrium, the heat gained by the mug must equal the heat lost by the tea:
Q_mug = -Q_tea
Solving these equations will give us the final temperature of the tea.
Let's substitute the given values and solve for T_final:
mcΔT_mug = -mcΔT_tea
(125 g)(4.18 J/goC)(T_final - 24.0oC) = -(341 g)(4.18 J/goC)(80.0oC - T_final)
Simplifying the equation:
523.75(T_final - 24.0) = -1425.14(80 - T_final)
523.75T_final - 12540 = -114011.2 + 1425.14T_final
902.39T_final = 101551.2
T_final = 112.7oC
Therefore, the final temperature of the tea after it attains equilibrium with the aluminum mug is 112.7oC.