You decide to take a nice hot bath but discover that your thoughtless roommate has used up most of the hot water. You fill up 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) Calculate the final temperature of the bath water.
(b) Calculate the net change in entropy of the system (bath water + boiling water), assuming no heat exchange with the air or the tub itself.
sum of heats gained is zero.
heat in original water added + heat in boiling water gained is zero.
270c(Tf-30)+5c(100-Tf)=0
solve for Tf
To calculate the final temperature of the bath water, we can use the principle of conservation of energy. The total heat gained by the bath water is equal to the total heat lost by the boiling water.
(a) Calculation of final temperature:
The specific heat capacity of water is 4186 J/kg°C.
The heat gained by the bath water is given by:
Q1 = m1 * c1 * ΔT1
where,
m1 = mass of bath water = 270 kg
c1 = specific heat capacity of water = 4186 J/kg°C
ΔT1 = change in temperature = Tf - 30.0°C (since the initial temperature is 30.0°C)
The heat lost by the boiling water is given by:
Q2 = m2 * c2 * ΔT2
where,
m2 = mass of boiling water = 5.00 kg
c2 = specific heat capacity of water = 4186 J/kg°C
ΔT2 = change in temperature = 100°C
Since Q1 = -Q2 (heat gained by one equals heat lost by the other), we can write:
m1 * c1 * ΔT1 = -m2 * c2 * ΔT2
Simplifying, we have:
m1 * ΔT1 = -m2 * ΔT2
Substituting the values:
270 kg * (Tf - 30.0°C) = -5.00 kg * (Tf - 100°C)
Expanding the equation, we get:
270 Tf - 8100 = -5 Tf + 500
Combining like terms, we have:
275 Tf = 8600
Solving for Tf, we get:
Tf = 31.27°C
Therefore, the final temperature of the bath water is approximately 31.27°C.
(b) Calculation of net change in entropy:
The change in entropy (ΔS) can be calculated using the equation:
ΔS = ∆Q / T
where ∆Q is the heat transferred and T is the temperature in Kelvin.
The net change in entropy (ΔSnet) is the sum of the changes in entropy for the bath water (∆S1) and the boiling water (∆S2), since no heat exchange is assumed with the surrounding air or the tub.
∆S1 = ∆Q1 / T1
where,
∆Q1 = heat gained by the bath water = m1 * c1 * ∆T1
T1 = initial temperature of the bath water = 30.0°C = 303.15 K
∆S2 = ∆Q2 / T2
where,
∆Q2 = heat lost by the boiling water = -m2 * c2 * ∆T2
T2 = initial temperature of the boiling water = 100.0°C = 373.15 K
Substituting the values and calculating, we have:
∆S1 = (270 kg * 4186 J/kg°C * (31.27°C - 30.0°C)) / 303.15 K
∆S2 = (-5.00 kg * 4186 J/kg°C * (31.27°C - 100.0°C)) / 373.15 K
Calculating the values, we get:
∆S1 = 1515.71 J/K
∆S2 = -8801.99 J/K
Therefore, the net change in entropy (∆Snet) is the sum of the individual changes:
∆Snet = ∆S1 + ∆S2
∆Snet = 1515.71 J/K - 8801.99 J/K
∆Snet = -7286.28 J/K
(a) To calculate the final temperature of the bath water, you can use the principle of conservation of energy. The heat gained by the bath water should be equal to the heat lost by the boiling water.
First, let's calculate the heat gained by the bath water:
Q1 = mcΔT1
where Q1 is the heat gained, m is the mass of the bath water, c is the specific heat capacity of water, and ΔT1 is the change in temperature of the bath water.
Q1 = (270 kg)(4.184 J/g°C)(Tf - 30.0°C)
Next, let's calculate the heat lost by the boiling water:
Q2 = mcΔT2
where Q2 is the heat lost, 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.
Q2 = (5.00 kg)(4.184 J/g°C)(100.0°C - Tf)
According to the principle of conservation of energy, Q1 = Q2.
Therefore, we can set the two equations equal to each other:
(270 kg)(4.184 J/g°C)(Tf - 30.0°C) = (5.00 kg)(4.184 J/g°C)(100.0°C - Tf)
Solving for Tf will give us the final temperature of the bath water.
(b) To calculate the net change in entropy of the system, we need to consider the entropic change of both the bath water and the boiling water.
The change in entropy can be calculated using the formula:
ΔS = Q/T
where ΔS is the change in entropy, Q is the heat transferred, and T is the temperature.
For the bath water, we can use Q1 as the heat transferred and the final temperature Tf as the temperature.
ΔS1 = Q1/Tf
For the boiling water, we can use Q2 as the heat transferred and the initial temperature (100.0°C) as the temperature.
ΔS2 = Q2/Tboiling (boiling point of water)
The net change in entropy is given by the sum of the individual changes in entropy:
ΔStotal = ΔS1 + ΔS2
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