A rock of mass 0.234 kg falls from rest from a height of 28.9 m into a pail containing 0.492 kg of water. The rock and water have the same initial temperature. The specific heat capacity of the rock is 1840 J/kg C°. Ignore the heat absorbed by the pail itself, and determine the rise in temperature of the rock and water in Celsius degrees
How do I do this ?
Qlost = Qgain??
To solve this problem, we need to apply the principle of energy conservation, which states that the heat lost by one object must be equal to the heat gained by another object.
In this case, we have a rock falling into a pail of water, and we want to find the rise in temperature of both the rock and the water.
Let's break down the steps:
1. Calculate the potential energy (PE) of the rock at its initial position using the formula PE = mgh, where m is the mass of the rock (0.234 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height (28.9 m).
PE = 0.234 kg * 9.8 m/s^2 * 28.9 m
2. The potential energy is converted into heat energy as the rock falls. This heat energy can be calculated using the formula Qgain = m * c * ΔT, where m is the mass of the rock (0.234 kg), c is the specific heat capacity of the rock (1840 J/kg°C), and ΔT is the change in temperature of the rock.
Qgain = 0.234 kg * 1840 J/kg°C * ΔT
3. The water in the pail will absorb the heat gained by the rock, so the heat gained by the water is equal to the heat lost by the rock.
Qlost = Qgain
4. The heat lost by the rock can be calculated using the formula Qlost = m * c * ΔT, where m is the mass of the water (0.492 kg), c is the specific heat capacity of water (4186 J/kg°C), and ΔT is the change in temperature of the water.
Qlost = 0.492 kg * 4186 J/kg°C * ΔT
5. Set Qlost equal to Qgain and solve for ΔT.
0.492 kg * 4186 J/kg°C * ΔT = 0.234 kg * 1840 J/kg°C * ΔT
Now, you can solve for ΔT, which will give you the rise in temperature of both the rock and the water.