A hot water tap of a bath delivers water at 85⁰C at a rate of 8Kg per minute. The cold water tap at the bath delivers water at 30⁰C at the rate of 20kg per minutes. If both tapes are at a time of 5 minutes. Calculate the final temperature o f the bath water in kelvin's, ignoring heat loses to the surrounding.

Answer 1

To calculate the final temperature of the bath water, we can use the principle of conservation of energy.

The heat gained by the hot water = (mass)(specific heat capacity)(change in temperature)
The heat gained by the cold water = (mass)(specific heat capacity)(change in temperature)
The total heat gained = heat gained by hot water + heat gained by cold water

Let's assume the specific heat capacity of water is 4200 J/kg°C.

Heat gained by hot water = (8kg)(4200 J/kg°C)(85-30) = 1,344,000 J
Heat gained by cold water = (20kg)(4200 J/kg°C)(T-30) where T is the final temperature in °C
Total heat gained = 1,344,000 J + (20kg)(4200 J/kg°C)(T-30)

Since the total heat gained is conserved:

1,344,000 J + (20kg)(4200 J/kg°C)(T-30) = 0 J
20(4200)(T-30) = -1,344,000
84000T - 1,260,000 = -1,344,000
84000T = -1,344,000 + 1,260,000
84000T = 84,000
T = 84,000 / 84,000
T = 1°C

So the final temperature of the bath water in °C is 1°C. To convert it to Kelvin, we add 273.15.

Final temperature in Kelvin = 1°C + 273.15 = 274.15 K

Therefore, the final temperature of the bath water in Kelvin is 274.15 K.

Answer 2

surely you can see that your answer is ridiculous.
Do you really think that mixing water at 85⁰C with water at 30⁰C will produce water at 1⁰C?

Answer 3

You are correct, and I apologize for the mistake in my previous calculation. Let's correct that calculation.

Let T represent the final temperature of the bath water in Celsius.

The total heat gained is conserved, so we can set up the equation:

(8 kg * 4200 J/kg°C * (85 - T)) + (20 kg * 4200 J/kg°C * (T - 30)) = 0

Simplifying the equation:

33600(85 - T) + 84000(T - 30) = 0
2856000 - 33600T + 84000T - 2520000 = 0
50400T = 336000
T = 336000 / 50400
T = 6.67°C

Now, to convert this to Kelvin:

Final temperature in Kelvin = 6.67°C + 273.15 = 279.82 K

Therefore, the final temperature of the bath water in Kelvin is approximately 279.82 K. Thank you for pointing out the error, and I appreciate your understanding.

Answer 4

6.67⁰C is no better.
Surely the mixture will have a temperature between 30⁰C and 85⁰C.

Answer 5

You are absolutely correct, and I appreciate your patience. Let's calculate the final temperature of the bath water more accurately this time.

Using the principle of conservation of energy, the heat lost by the hot water equals the heat gained by the cold water:

(Mass hot water * Specific heat capacity * (Initial temperature hot water - Final temperature)) = (Mass cold water * Specific heat capacity * (Final temperature - Initial temperature cold water))

(8kg * 4200J/kg°C * (85°C - T)) = (20kg * 4200J/kg°C * (T - 30°C))

33600(85 - T) = 84000(T - 30)
2856000 - 33600T = 84000T - 2520000
2856000 + 2520000 = 84000T + 33600T
5376000 = 117600T
T = 5376000 / 117600
T = 45.71°C

Converting this to Kelvin:
Final temperature in Kelvin = 45.71°C + 273.15
Final temperature in Kelvin = 318.86 K

Therefore, the final temperature of the bath water will be approximately 318.86 K when hot water at 85°C and cold water at 30°C are mixed in the given quantities. Thank you for your understanding, and I hope this answer is more accurate.