a hot copper of mass 2 Kg is at a temperature of 150 degrees celcius . 0.1 Kg of cool water at 25 degrees celcius, is poured into the pot. The lid is replaced quickly to ensure no steam can escape. Assume thta no heat is lost to the surroundings, find the final temperature of the pot and its contents
To find the final temperature of the pot and its contents, we need to apply the principle of conservation of energy. The heat lost by the hot copper pot is equal to the heat gained by the cool water.
The equation for heat transfer is given by:
Q = m * c * ΔT
Where:
Q = Heat transfer (in Joules)
m = Mass (in kilograms)
c = Specific heat capacity (in J/kg°C)
ΔT = Change in temperature (in °C)
For the copper pot, using the equation Q = m * c * ΔT:
Q1 = (2 kg) * (specific heat capacity of copper) * (final temperature - 150°C)
For the water, using the equation Q = m * c * ΔT:
Q2 = (0.1 kg) * (specific heat capacity of water) * (final temperature - 25°C)
Since the heat lost by the pot is equal to the heat gained by the water, we can equate Q1 and Q2:
(2 kg) * (specific heat capacity of copper) * (final temperature - 150°C) = (0.1 kg) * (specific heat capacity of water) * (final temperature - 25°C)
Simplifying the equation:
2 * (final temperature - 150) = 0.1 * (final temperature - 25)
Expanding and re-arranging the equation:
2 * final temperature - 300 = 0.1 * final temperature - 2.5
1.9 * final temperature = 297.5
final temperature = 297.5 / 1.9
final temperature ≈ 156.58°C
Therefore, the final temperature of the pot and its contents would be approximately 156.58 degrees Celsius.