At 28.4 g sample of aluminum is heated to 39.4 degrees Celsius and then is placed in a calorimeter containing 50.0g of water. Temp of water increases from 21.00 degrees Celsius to 23.00 degrees Celsius. What is the specific heat of aluminum?
the sum of heats gained is zero.
HeatgainedAl+heatgainedwater=0
28.4*cAl*(23-39.4)+50*cWater*(23-21)=0
solve for cAl
To find the specific heat of aluminum, we can use the formula:
Q = m * c * ΔT
Where:
Q = heat gained or lost
m = mass of the substance
c = specific heat capacity of the substance
ΔT = change in temperature
In this case, the aluminum sample is heated, so it gains heat. The water within the calorimeter absorbs this heat, causing its temperature to increase. We need to find the specific heat capacity of aluminum, represented by c.
First, let's calculate the heat gained or lost by the water:
Q_water = m_water * c_water * ΔT_water
Where:
m_water = mass of water
c_water = specific heat capacity of water
ΔT_water = change in temperature of water
Given values:
m_water = 50.0g
c_water = 4.18 J/g°C (specific heat capacity of water)
ΔT_water = 23.00°C - 21.00°C = 2.00°C
Plugging these values into the equation, we get:
Q_water = 50.0g * 4.18 J/g°C * 2.00°C
Now, let's calculate the heat gained or lost by the aluminum:
Q_aluminum = m_aluminum * c_aluminum * ΔT_aluminum
Where:
m_aluminum = mass of aluminum
c_aluminum = specific heat capacity of aluminum
ΔT_aluminum = change in temperature of aluminum
Given values:
m_aluminum = 28.4g
c_aluminum = ? (what we need to find)
ΔT_aluminum = 39.4°C - initial temperature of the aluminum (unknown)
To find ΔT_aluminum, we can use the fact that the heat gained by the aluminum is equal to the heat lost by the water. Therefore:
Q_water = Q_aluminum
Plugging in the values we know, we have:
50.0g * 4.18 J/g°C * 2.00°C = 28.4g * c_aluminum * ΔT_aluminum
Now we can solve for c_aluminum:
c_aluminum = (50.0g * 4.18 J/g°C * 2.00°C) / (28.4g * ΔT_aluminum)
To find ΔT_aluminum, subtract the initial temperature of the aluminum from the final temperature of the water:
ΔT_aluminum = 23.00°C - initial temperature of the aluminum
Finally, substitute this value back into the equation to find c_aluminum.