For a bathing person supplies 5×10^3Kj energy in 20 litre of water at 15°c temperature.What is the change in temperature of this water?
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To raise 1 gram of water 1 degree Centigrade, it takes approximately 4.184 Joules of energy.
In your 20 Litres, you have about 20,000 grams of water.
The energy you have available for use in your problem is 5000 kJ or 5,000,000 joules.
(5,000,000 joules) / [(20,000 grams)*(4.184 joules per gram per degree)] = 59.8 degree increase in water temperature.
Which means that the bath water would be almost 75 degrees C - way too hot!
To find the change in temperature of the water, we need to use the specific heat equation:
Q = mcΔT
Where:
Q is the amount of energy transferred to the water
m is the mass of the water
c is the specific heat capacity of water, which is approximately 4.18 J/g°C
ΔT is the change in temperature
First, we need to convert the energy from kilojoules to joules. As 1 kJ = 1000 J, the energy supplied is 5 × 10^3 kJ = 5 × 10^6 J.
Next, we need to determine the mass of the water. Given that the volume is 20 liters and the density of water is approximately 1 gram per milliliter, the mass can be calculated as:
mass = volume × density
mass = 20 L × 1000 mL/L × 1 g/mL
mass = 20000 g
Now we can substitute the values into the specific heat equation to find the change in temperature:
Q = mcΔT
5 × 10^6 J = 20000 g × 4.18 J/g°C × ΔT
Simplifying the equation gives:
ΔT = (5 × 10^6 J) / (20000 g × 4.18 J/g°C)
ΔT ≈ 59.33°C
Therefore, the change in temperature of the water is approximately 59.33°C.