The temperature of 100.0g of water initially at 25.0 degrees C in a calorimeter increases to 30.0 degrees C after adding a 20.0g piece of hot aluminum. What was the temperature of the aluminum before adding it to water? The specific heat of water and aluminum is 4.182J/g and .900J/g, respectively.
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m*C*DeltaT of Water = m*C*DeltaT of Al
Al was 86 degrees C
To solve this problem, we can use the equation for heat transfer:
q = mcΔT
Where:
q = heat transfer (in joules)
m = mass (in grams)
c = specific heat capacity (in joules per gram-degree Celsius)
ΔT = change in temperature (in degrees Celsius)
We will use this equation for both the water and the aluminum.
For water:
q_water = (100.0g) * (4.182 J/g°C) * (30.0°C - 25.0°C)
For aluminum:
q_aluminum = (20.0g) * (0.900 J/g°C) * (T - 25.0°C)
Now, since there is no heat lost or gained in this system (assuming it is isolated), we can set the heat gained by the water equal to the heat lost by the aluminum:
q_water = q_aluminum
Substituting the values:
(100.0g) * (4.182 J/g°C) * (30.0°C - 25.0°C) = (20.0g) * (0.900 J/g°C) * (T - 25.0°C)
Simplifying:
(100.0g) * (4.182 J/g°C) * 5.0°C = (20.0g) * (0.900 J/g°C) * (T - 25.0°C)
20.91 J/°C = (0.9 J/g°C) * (T - 25.0°C)
Dividing both sides by 0.9 J/g°C:
23.23 °C = T - 25.0°C
Simplifying:
T - 25.0°C = 23.23°C
Adding 25.0 °C to both sides:
T = 48.23°C
Therefore, the temperature of the aluminum before adding it to the water was approximately 48.23°C.
To determine the temperature of the aluminum before adding it to water, we can use the principle of heat transfer. The heat lost by the aluminum is equal to the heat gained by the water and the calorimeter.
First, let's calculate the heat gained by the water and the calorimeter. We can use the formula:
Q = mcΔT
Where:
Q is the heat gained (or lost),
m is the mass of the object (water and calorimeter),
c is the specific heat capacity of the object, and
ΔT is the change in temperature.
For the water and calorimeter, we have:
m = 100.0g (mass of water) + mass of calorimeter
c = 4.182 J/g (specific heat of water)
ΔT = (30.0 degrees C) - (25.0 degrees C) = 5.0 degrees C
Using the formula, we can calculate the heat gained by the water and calorimeter:
Q_water_calorimeter = (100.0g + mass of calorimeter) * 4.182 J/g * 5.0 degrees C
Next, let's calculate the heat lost by the aluminum. Similarly, we can use the formula:
Q = mcΔT
For the aluminum, we have:
m = 20.0g (mass of aluminum)
c = 0.900 J/g (specific heat of aluminum)
ΔT = (30.0 degrees C) - (T_aluminum) (change in temperature)
Using the formula, we can calculate the heat lost by the aluminum:
Q_aluminum = 20.0g * 0.900 J/g * (30.0 degrees C - T_aluminum)
Since the heat lost by the aluminum is equal to the heat gained by the water and calorimeter, we can set up an equation:
Q_water_calorimeter = Q_aluminum
(100.0g + mass of calorimeter) * 4.182 J/g * 5.0 degrees C = 20.0g * 0.900 J/g * (30.0 degrees C - T_aluminum)
Now, we can solve for T_aluminum:
(100.0g + mass of calorimeter) * 4.182 J/g * 5.0 degrees C = 20.0g * 0.900 J/g * (30.0 degrees C - T_aluminum)
Simplifying the equation:
(100.0g + mass of calorimeter) * 4.182 J/g * 5.0 degrees C = 18.0g * (30.0 degrees C - T_aluminum)
Now, we need to know the mass of the calorimeter in order to solve for T_aluminum. The mass of the calorimeter is not provided in the problem statement.