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.
This looks like a duplicate post. Look for the answer in the thread of one of your other posts.
We did not find results for: a typical woman gives off heat at rate of about 8000 calories per hour. how long would a woman have to stay in a bath of 60L (60,000 g) of 26 C water in order to raise the water temperature to 30 c ? presure that all the heat given off by the woman is transferred to the water , and that the water does not lose any heat to the air. show your calculations..
To answer these questions, we can use the principle of conservation of energy and the specific heat capacity equation.
(a) Whether this process is reversible or irreversible can be determined by considering the assumption of no heat exchange with the air or tub. In a reversible process, the system and surroundings can be returned to their original states with no net change. However, in this situation, it is unlikely that the bath water and boiling water can be returned to their original states without heat exchange with the air or tub. Therefore, it can be concluded that this process is irreversible.
(b) To calculate the final temperature of the bath water, we can use the principle of conservation of energy. In this case, the heat gained by the bath water is equal to the heat lost by the boiling water. The equation for heat transfer can be written as:
m1 * c1 * ΔT1 = m2 * c2 * ΔT2
where:
m1 = mass of the bath water (270 kg)
c1 = specific heat capacity of water (4.186 J/g°C)
ΔT1 = change in temperature of the bath water (final temperature - initial temperature)
m2 = mass of the boiling water (5.00 kg)
c2 = specific heat capacity of water (4.186 J/g°C)
ΔT2 = change in temperature of the boiling water (final temperature - initial temperature)
Since the boiling water is initially at its boiling point, which is 100°C, and the bath water is initially at 30.0°C, the equation becomes:
270 kg * 4.186 J/g°C * (Tf - 30.0°C) = 5.00 kg * 4.186 J/g°C * (Tf - 100°C)
Simplifying the equation:
1131.42 kg * Tf - 33942.6 J = 20.93 kg * Tf - 20930 J
Subtracting 20.93 kg * Tf from both sides:
(1131.42 kg - 20.93 kg) * Tf - 33942.6 J = -20930 J
Further simplification:
1110.49 kg * Tf - 33942.6 J = -20930 J
Adding 33942.6 J to both sides:
1110.49 kg * Tf = 13012.6 J
Dividing both sides by 1110.49 kg:
Tf ≈ 11.7°C
Therefore, the final temperature of the bath water is approximately 11.7°C.
(c) To calculate the net change in entropy of the system, we can use the formula:
ΔS = q / T
where:
ΔS = change in entropy
q = heat transferred
T = temperature in Kelvin
Since there is no heat exchange with the air or tub, the bath water and boiling water are the only components of the system. The heat transferred, in this case, is the heat gained by the bath water. Using the specific heat capacity equation, we can determine the heat transferred:
q = m1 * c1 * ΔT1
where:
m1 = mass of the bath water (270 kg)
c1 = specific heat capacity of water (4.186 J/g°C)
ΔT1 = change in temperature of the bath water (final temperature - initial temperature)
Substituting the given values:
q = 270 kg * 4.186 J/g°C * (11.7°C - 30.0°C)
Calculating this gives:
q = -39925.2 J
Now, we can calculate the net change in entropy:
ΔS = -39925.2 J / (11.7°C + 273.15 K) [converting to Kelvin]
ΔS ≈ -124 J/K
Therefore, the net change in entropy of the system is approximately -124 J/K.