An aluminum pan of mass 100 grams and at a temperature of 375°C is plunged into a vat of water.
a. If the pan and the water end up at 45°C, how much energy did the aluminum pan lose?
b. Assuming no energy goes into the environment, how much energy must the water have gained from the aluminum?
a.
q = mass Al x specific heat Al x (Tfinal-Tinitial)
b.
Look at your specific heat problem just before this as a hint (if you need a hint).
To determine the energy lost by the aluminum pan and gained by the water, we can use the heat transfer equation:
Q = mcΔT
Where:
Q is the heat energy exchanged
m is the mass of the substance
c is the specific heat capacity of the substance
ΔT is the change in temperature
a. To calculate the energy lost by the aluminum pan:
1. Determine the specific heat capacity of aluminum. The specific heat capacity of aluminum is 0.897 J/g·°C.
2. Calculate the change in temperature experienced by the aluminum pan:
ΔT = final temperature - initial temperature
Given:
Initial temperature of the aluminum pan = 375°C
Final temperature of the aluminum pan = 45°C
ΔT = 45°C - 375°C = -330°C
Note: ΔT is negative because the temperature of the aluminum pan decreased.
3. Calculate the energy lost:
Q = (mass of the aluminum pan) × (specific heat capacity of aluminum) × (change in temperature)
Given:
Mass of the aluminum pan = 100 grams
Q = 100 g × 0.897 J/g·°C × -330°C
b. To calculate the energy gained by the water:
1. Determine the specific heat capacity of water. The specific heat capacity of water is 4.18 J/g·°C.
2. Calculate the change in temperature experienced by the water:
ΔT = final temperature - initial temperature
Given:
Initial temperature of the water = 45°C
Final temperature of the water = 45°C (since it is stated that the water and the pan end up at the same temperature)
ΔT = 45°C - 45°C = 0°C
Note: ΔT is zero because there is no change in temperature.
3. Calculate the energy gained:
Q = (mass of the water) × (specific heat capacity of water) × (change in temperature)
Given:
Mass of the water = mass of the aluminum pan (since both are plunged together)
Q = 100 g × 4.18 J/g·°C × 0°C
Now, you can calculate the energy lost by the aluminum pan (Qa) and the energy gained by the water (Qw) using the above equations.