In a calorimetry experiment, an object of mass M= 9.2 kg falls through a height of 35.5 m and, by means of a mechanical linkage, rotates a paddle wheel that stirs m= 0.92 kg of water. The water is initially at 20∘C. What is the maximum possible temperature rise of the water? (The specific heat of water is 4186J/kg⋅K).
To find the maximum possible temperature rise of the water, we need to calculate the amount of thermal energy transferred from the falling object to the water.
1. Calculate the gravitational potential energy (GPE) of the object:
GPE = mgh
where m is the mass (9.2 kg) and h is the height (35.5 m).
GPE = 9.2 kg * 9.8 m/s^2 * 35.5 m
= 3,180.76 J
2. Since the paddle wheel is linked mechanically to the falling object, all the GPE is converted to work done on the water. Therefore, the work done on the water is also 3,180.76 J.
3. The work done on the water is given by:
Work = force × distance
The force can be calculated using:
Force = mass × acceleration
The acceleration can be calculated using:
acceleration = (final velocity - initial velocity) / time
In this case, the object is initially at rest and falls freely, so the initial velocity is 0 and the acceleration is due to gravity (9.8 m/s^2).
acceleration = 9.8 m/s^2
The final velocity can be calculated using:
final velocity = sqrt(2gh)
final velocity = sqrt(2 * 9.8 m/s^2 * 35.5 m)
= 59.0421 m/s
The time taken for the object to fall can be calculated using:
time = distance / average velocity
The average velocity can be calculated using:
average velocity = (final velocity + initial velocity) / 2 = final velocity / 2
time = 35.5 m / (59.0421 m/s) = 0.6021 s
The force can be calculated as:
Force = mass × acceleration = 0.92 kg * 9.8 m/s^2 = 9.056 N
Now we can calculate the work done on the water:
Work = Force × distance = 9.056 N × 35.5 m = 321.968 J
4. The thermal energy transferred to the water is equal to the work done on the water. Since there is no other energy loss in the system, we can assume that all the thermal energy transferred is used to heat the water.
Thermal energy transferred = 321.968 J
5. To calculate the temperature rise of the water, we can use the equation:
Q = m × c × ΔT
Where Q is the thermal energy transferred (321.968 J), m is the mass of the water (0.92 kg), c is the specific heat of water (4186 J/kg·K), and ΔT is the change in temperature.
Rearranging the equation to solve for ΔT:
ΔT = Q / (m × c)
ΔT = 321.968 J / (0.92 kg × 4186 J/kg·K)
≈ 0.081 K
Therefore, the maximum possible temperature rise of the water is approximately 0.081 K.
To find the maximum possible temperature rise of the water in the calorimetry experiment, we need to calculate the amount of energy transferred from the falling object to the water.
First, let's find the potential energy (PE) of the falling object. PE is given by the formula:
PE = mgh
Where:
m = mass of the falling object (9.2 kg)
g = acceleration due to gravity (approximately 9.8 m/s²)
h = height (35.5 m)
Substituting the given values:
PE = 9.2 kg * 9.8 m/s² * 35.5 m
= 3160.76 kg·m²/s² or Joules (J)
Next, we need to determine the amount of energy transferred to the water by rotating the paddle wheel. This is equal to the work done on the paddle wheel. The work done (W) is given by:
W = force * distance
In this case, the force exerted on the paddle wheel by the falling object is equal to its weight. The weight (W) is given by:
W = mass * gravity
W = 9.2 kg * 9.8 m/s²
= 90.16 N
The distance through which the force is exerted is the circumference of the paddle wheel. Let's assume the paddle wheel has a radius of r meters. So, the distance (d) is given by:
d = 2πr
We don't have the radius of the paddle wheel, so we cannot calculate the exact distance. However, we know that the distance the object falls (35.5 m) is equal to the distance the paddle wheel rotates. Therefore, we can write:
d = 35.5 m
Now we can calculate the work done on the paddle wheel:
W = 90.16 N * 35.5 m
= 3203.48 N·m or Joules (J)
The energy transferred to the water is equal to the work done on the paddle wheel. So, the energy transfer is 3203.48 J.
Finally, to calculate the temperature rise of the water, we can use the formula:
Q = mcΔT
Where:
Q = energy transfer (3203.48 J)
m = mass of the water (0.92 kg)
c = specific heat capacity of water (4186 J/kg·K)
ΔT = temperature rise
Rearranging the formula to solve for ΔT:
ΔT = Q / (mc)
Substituting the given values:
ΔT = (3203.48 J) / (0.92 kg * 4186 J/kg·K)
≈ 0.85 K
Therefore, the maximum possible temperature rise of the water in the calorimetry experiment is approximately 0.85 degrees Celsius.