In a Joule experiment, a mass of 1.00 kg falls through a height of 50.0 m and rotates a paddle wheel that stirs 0.600 kg of water. The water is initially at 15 degrees C. By how much does its temperature rise (in degrees C)?
To find out how much the temperature of the water rises in this Joule experiment, we need to use the principle of conservation of energy. The gravitational potential energy of the falling mass is converted into both kinetic energy and thermal energy.
First, let's calculate the gravitational potential energy (GPE) of the 1.00 kg mass as it falls through a height of 50.0 m. The formula for GPE is given by:
GPE = m * g * h
Where:
m = mass (1.00 kg)
g = acceleration due to gravity (9.8 m/s^2)
h = height (50.0 m)
Plugging in the values, we get:
GPE = 1.00 kg * 9.8 m/s^2 * 50.0 m
GPE = 490 J
Now, to calculate the thermal energy gained by the water, we need to know its specific heat capacity. The specific heat capacity of water is 4186 J/kg°C.
The formula for thermal energy is given by:
Thermal energy = mass * specific heat capacity * temperature change
Where:
mass = mass of water (0.600 kg)
specific heat capacity = specific heat capacity of water (4186 J/kg°C)
temperature change = final temperature - initial temperature
The final temperature is what we need to find. The initial temperature is given as 15°C.
Let's assume the final temperature is T°C.
Using the conservation of energy, we can set up the equation:
GPE = Thermal energy
Plugging in the values, we get:
490 J = 0.600 kg * 4186 J/kg°C * (T°C - 15°C)
Now, we can solve this equation for T°C.
Divide both sides by (0.600 kg * 4186 J/kg°C):
490 J / (0.600 kg * 4186 J/kg°C) = T°C - 15°C
Simplifying the left side:
1.707 = T°C - 15°C
Adding 15°C to both sides:
T°C = 1.707 + 15°C
T°C = 16.707°C
Therefore, the temperature of the water would rise by approximately 16.707°C in this Joule experiment.