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Equal masses of three liquids A, B and C have temperatures 10oC, 25oC and 40oC respectively. If A and B are mixed, the mixture has a temperature of 15oC. If B and C are mixed, the mixture has a temperature of 30oC,. If A and C are mixed the mixture will have a temperature of
- a)16oC
- b)20oC
- c)25oC
- d)29oC
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
Equal masses of three liquids A, B and C have temperatures 10oC, 25oC ...
msA (15 – 10) = msB (25 – 15) sA = 2sB
msB (30 – 25) = msC (40 – 30)
sB = 2sC ⇒ sA = 4sC
msA (T – 10) = msC (40 – T)
⇒ 4(T – 10) = 40 – T
T = 16°C
msB (30 – 25) = msC (40 – 30)
sB = 2sC ⇒ sA = 4sC
msA (T – 10) = msC (40 – T)
⇒ 4(T – 10) = 40 – T
T = 16°C
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Equal masses of three liquids A, B and C have temperatures 10oC, 25oC ...
Given:
- Mass of liquids A, B, and C is equal.
- Temperatures of A, B, and C are 10°C, 25°C, and 40°C respectively.
- When A and B are mixed, the mixture has a temperature of 15°C.
- When B and C are mixed, the mixture has a temperature of 30°C.
To find:
The temperature of the mixture when A and C are mixed.
Solution:
Let's assume the specific heat capacity of the liquids is the same and equal to 1 (for simplicity).
When A and B are mixed:
- The total heat gained by A and B is equal to the heat lost by the mixture.
- Mass of A = Mass of B = m (let's assume)
- Temperature of A = 10°C, Temperature of B = 25°C
- Temperature of the mixture = 15°C
- Heat lost by the mixture = (mass of A + mass of B) * specific heat capacity * (temperature of the mixture - initial temperature of the mixture)
- Heat lost by the mixture = 2m * 1 * (15 - 15) = 0
When B and C are mixed:
- The total heat gained by B and C is equal to the heat lost by the mixture.
- Mass of B = Mass of C = m (let's assume)
- Temperature of B = 25°C, Temperature of C = 40°C
- Temperature of the mixture = 30°C
- Heat lost by the mixture = (mass of B + mass of C) * specific heat capacity * (temperature of the mixture - initial temperature of the mixture)
- Heat lost by the mixture = 2m * 1 * (30 - 30) = 0
From the above two scenarios, we can conclude that when two liquids at different temperatures are mixed, the mixture temperature will be the average of the initial temperatures if the masses and specific heat capacities are equal.
Therefore, when A and C are mixed:
- The total heat gained by A and C is equal to the heat lost by the mixture.
- Mass of A = Mass of C = m (let's assume)
- Temperature of A = 10°C, Temperature of C = 40°C
- Let the temperature of the mixture be T.
- Heat lost by the mixture = (mass of A + mass of C) * specific heat capacity * (temperature of the mixture - initial temperature of the mixture)
- Heat lost by the mixture = 2m * 1 * (T - T) = 0
Since the heat lost by the mixture is 0, it implies that the temperature of the mixture will be the average of the initial temperatures of A and C.
Average temperature = (Temperature of A + Temperature of C) / 2 = (10 + 40) / 2 = 50 / 2 = 25°C
Therefore, the temperature of the mixture when A and C are mixed is 25°C.
Hence, the correct answer is option 'C'.
- Mass of liquids A, B, and C is equal.
- Temperatures of A, B, and C are 10°C, 25°C, and 40°C respectively.
- When A and B are mixed, the mixture has a temperature of 15°C.
- When B and C are mixed, the mixture has a temperature of 30°C.
To find:
The temperature of the mixture when A and C are mixed.
Solution:
Let's assume the specific heat capacity of the liquids is the same and equal to 1 (for simplicity).
When A and B are mixed:
- The total heat gained by A and B is equal to the heat lost by the mixture.
- Mass of A = Mass of B = m (let's assume)
- Temperature of A = 10°C, Temperature of B = 25°C
- Temperature of the mixture = 15°C
- Heat lost by the mixture = (mass of A + mass of B) * specific heat capacity * (temperature of the mixture - initial temperature of the mixture)
- Heat lost by the mixture = 2m * 1 * (15 - 15) = 0
When B and C are mixed:
- The total heat gained by B and C is equal to the heat lost by the mixture.
- Mass of B = Mass of C = m (let's assume)
- Temperature of B = 25°C, Temperature of C = 40°C
- Temperature of the mixture = 30°C
- Heat lost by the mixture = (mass of B + mass of C) * specific heat capacity * (temperature of the mixture - initial temperature of the mixture)
- Heat lost by the mixture = 2m * 1 * (30 - 30) = 0
From the above two scenarios, we can conclude that when two liquids at different temperatures are mixed, the mixture temperature will be the average of the initial temperatures if the masses and specific heat capacities are equal.
Therefore, when A and C are mixed:
- The total heat gained by A and C is equal to the heat lost by the mixture.
- Mass of A = Mass of C = m (let's assume)
- Temperature of A = 10°C, Temperature of C = 40°C
- Let the temperature of the mixture be T.
- Heat lost by the mixture = (mass of A + mass of C) * specific heat capacity * (temperature of the mixture - initial temperature of the mixture)
- Heat lost by the mixture = 2m * 1 * (T - T) = 0
Since the heat lost by the mixture is 0, it implies that the temperature of the mixture will be the average of the initial temperatures of A and C.
Average temperature = (Temperature of A + Temperature of C) / 2 = (10 + 40) / 2 = 50 / 2 = 25°C
Therefore, the temperature of the mixture when A and C are mixed is 25°C.
Hence, the correct answer is option 'C'.
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Equal masses of three liquids A, B and C have temperatures 10oC, 25oC and 40oC respectively. If A and B are mixed, the mixture has a temperature of 15oC. If B and C are mixed, the mixture has a temperature of 30oC,. If A and C are mixed the mixture will have a temperature ofa)16oCb)20oCc)25oCd)29oCCorrect answer is option 'A'. Can you explain this answer?
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