Please show work
You decide to take a nice hot bath but discover that your thoughtless roommate has used up most of the hot water. You fill the tub with 270 kg of 30.0°C water and attempt to warm it further by pouring in 5.00 kg of boiling water from the stove.
(a) Is this a reversible or an irreversible process? Use physical reasoning to explain.
(b) Calculate the final temperature of the bath water.
(c) Calculate the net change in entropy of the system (bath water + boiling water), assuming no heat exchange with the air or the tub itself.
a) Mixing high and low temperature is an irreversible process. You cannot separate the equilibrated mixture into high and low temperature without performing work. Could you run the mixing process backwards? No way.
b) Set the heat loss by the warm water equal to the heat gained by the cold water, and solve for final T.
c) Entropy gain =
Q/Tfinal - Q/t1 - Q/T2.
Get Q (the heat transfer) from part (b)
Oh, a hot water problem! Let's dive in!
(a) Is this a reversible or an irreversible process? Reversible means that the process can be reversed to return the system to its initial state. In this case, the process is irreversible because you can't really un-mix hot and cold water once they're mixed. Unless you have magical powers, of course!
(b) To calculate the final temperature, we can use the conservation of energy principle. The total heat gained by the cold water should be equal to the total heat lost by the hot water. So, we can write the equation as:
m₁c₁ΔT₁ = m₂c₂ΔT₂
Where m₁ and c₁ are the mass and specific heat capacity of the cold water, ΔT₁ is the change in temperature of the cold water, m₂ and c₂ are the mass and specific heat capacity of the hot water, and ΔT₂ is the change in temperature of the hot water.
Plugging in the values:
270 kg * specific heat of water * (final temperature - 30°C) = 5 kg * specific heat of water * (100°C - final temperature)
Solve this equation to find the final temperature. The specific heat capacity of water is approximately 4.18 J/g°C.
(c) To calculate the net change in entropy, we need to consider the entropy change of both the hot and cold water. Entropy change is given by the equation:
ΔS = Q/T
Where ΔS is the change in entropy, Q is the heat transfer, and T is the temperature.
Now, since there is no heat exchange with the air or the tub, Q will be the heat gained or lost by the water. We can calculate the entropy change of each water sample using their respective masses and temperatures.
Then, we can sum up the entropy changes of both the hot and cold water to find the net change in entropy of the system.
And there you have it! A hot bath problem with some bonus physics thrown in.
To answer these questions, we can use the law of conservation of energy and the equation for heat transfer:
1. Is this a reversible or an irreversible process?
In this case, the process is irreversible. It is because the hot water is being poured into the bathwater, and there is no way to recover the initial state of the system. Additionally, there may be heat loss to the surroundings during the pouring process.
2. Calculate the final temperature of the bath water.
To calculate the final temperature, we can use the principle of conservation of energy. The heat lost by the boiling water is equal to the heat gained by the bathwater.
The heat gained by the bathwater is given by:
q1 = mcΔT1,
where q1 is the heat gained by the bathwater, m is the mass of the bathwater, c is the specific heat capacity of water, and ΔT1 is the change in temperature of the bathwater.
The heat lost by the boiling water is given by:
q2 = mcΔT2,
where q2 is the heat lost by the boiling water, m is the mass of the boiling water, c is the specific heat capacity of water, and ΔT2 is the change in temperature of the boiling water.
Since there is no heat exchange with the air or the tub itself, the heat lost by the boiling water is equal to the heat gained by the bathwater:
q2 = q1
Therefore, we can write:
mcΔT2 = mcΔT1
Simplifying and rearranging the equation, we get:
ΔT2 = ΔT1
In this case, we have ΔT1 = final temperature - 30.0°C and ΔT2 = 100.0°C - final temperature.
Equating the two expressions for ΔT, we can solve for the final temperature:
final temperature - 30.0°C = 100.0°C - final temperature
Combining like terms:
2 * final temperature = 130.0°C
Simplifying:
final temperature = 65.0°C
Therefore, the final temperature of the bath water is 65.0°C.
3. Calculate the net change in entropy of the system (bath water + boiling water), assuming no heat exchange with the air or the tub itself.
The change in entropy (ΔS) of a system can be calculated using the equation:
ΔS = q / T,
where ΔS is the change in entropy, q is the heat transfer, and T is the temperature.
Since no heat exchange occurs with the air or the tub, the total change in entropy is due to heat transfer between the bath water and the boiling water.
The heat transfer is given by:
q = mcΔT
Substituting the values we know, and considering the mass and the specific heat capacity of water are the same for both the bath water and boiling water, we can write:
q = 270 kg * c * (final temperature - 30.0°C) + 5.00 kg * c * (final temperature - 100.0°C)
Substituting the specific heat capacity of water (c ≈ 4.18 J/g°C), we can calculate the net change in entropy.
Remember to convert kg to grams for the mass to match the units of the specific heat capacity (J/g°C).
I hope this helps! Let me know if you have any further questions.
To find the final temperature of the bath water and the net change in entropy of the system, we need to consider the principle of energy conservation and the law of entropy.
(a) Reversibility:
Determining whether a process is reversible or irreversible requires analyzing whether the system and its surroundings can return to their initial states without any changes. In this case, when you pour the boiling water into the bath, heat transfer occurs between the hot and cold water, making it an irreversible process. Once mixed, it is impossible to separate the hot and cold water and return them to their original states.
(b) Final temperature of the bath water:
To calculate the final temperature, we can apply the principle of energy conservation, which states that the total energy of an isolated system remains constant.
The heat gained by the cold water = heat lost by the hot water.
The heat gained by the cold water can be calculated using the equation:
Q(cold water) = mcΔT
Where:
m = mass of cold water = 270 kg
c = specific heat capacity of water = 4.18 kJ/(kg·°C)
ΔT = change in temperature = final temperature - initial temperature
initial temperature = 30.0 °C
The heat lost by the hot water can also be calculated using the equation:
Q(hot water) = mhΔT
Where:
m = mass of hot water = 5.00 kg
h = specific heat capacity of water = 4.18 kJ/(kg·°C)
ΔT = change in temperature = final temperature - boiling temperature
boiling temperature = 100 °C
Since the total energy remains constant, we can equate these two equations:
mcΔT = mhΔT
Substituting the known values:
270 kg * 4.18 kJ/(kg·°C) * (final temperature - 30.0 °C) = 5.00 kg * 4.18 kJ/(kg·°C) * (final temperature - 100 °C)
Simplifying the equation:
270 * (final temperature - 30.0) = 5.00 * (final temperature - 100)
Solve for the final temperature:
270 * final temperature - 270 * 30.0 = 5.00 * final temperature - 5.00 * 100
270 * final temperature - 8100 = 5.00 * final temperature - 500
265 * final temperature = 7600
final temperature ≈ 28.68 °C
Therefore, the final temperature of the bath water is approximately 28.68 °C.
(c) Net change in entropy of the system:
The net change in entropy of the system can be calculated using the equation:
ΔS(system) = Q(rev) / T
Where:
ΔS(system) = net change in entropy of the system
Q(rev) = heat transfer in a reversible process
T = temperature
Since the transfer of heat between the hot and cold water is irreversible, we cannot directly use this equation. However, we can still calculate the net change in entropy by considering the positive and negative contributions.
The positive contribution to entropy is due to the heat gained by the cold water:
ΔS(cold water) = Q(cold water) / T(cold water)
Where:
Q(cold water) = mcΔT (calculated above)
T(cold water) = final temperature of the bath water
Substituting the known values:
ΔS(cold water) = (270 kg * 4.18 kJ/(kg·°C) * (final temperature - 30.0 °C)) / (final temperature)
The negative contribution to entropy is from the heat lost by the hot water:
ΔS(hot water) = -Q(hot water) / T(hot water)
Where:
Q(hot water) = mhΔT (calculated above)
T(hot water) = boiling temperature
Substituting the known values:
ΔS(hot water) = -(5.00 kg * 4.18 kJ/(kg·°C) * (final temperature - 100 °C)) / (100 °C)
The net change in entropy of the system is the sum of these contributions:
ΔS(system) = ΔS(cold water) + ΔS(hot water)
Calculating:
ΔS(cold water) = (270 * 4.18 * (28.68 - 30.0)) / (28.68) ≈ -9.34 kJ/K
ΔS(hot water) = -(5.00 * 4.18 * (28.68 - 100)) / (100) ≈ 29.57 kJ/K
ΔS(system) = -9.34 + 29.57 ≈ 20.23 kJ/K
Therefore, the net change in entropy of the system is approximately 20.23 kJ/K.