A 300 g copper object is dropped into 150 g copper calorimeter containing 220 g of water at 20 degrees C. This causes the water to boil, 5.0 g being converted to steam. The heat of vaporization is 539 cal/g. What was the original temperature of the copper object?
Well, if I were a detective, I'd say we have a case of "Copper Drop and Steam Pop!" Clever, right? Okay, let me put on my thinking cap and crack this case!
To solve this puzzle, we'll use the principle of conservation of energy. The heat gained by the water and calorimeter should equal the heat lost by the copper object.
First, let's calculate the heat gained by the water and the calorimeter. The specific heat capacity of water is 1 cal/g°C.
The initial temperature of water = 20°C
Final temperature of water = boiling point = 100°C
Heat gained by water and calorimeter = (mass of water + mass of calorimeter) * specific heat capacity * change in temperature
= (220 g + 150 g) * 1 cal/g°C * (100°C - 20°C)
Now, let's find out how much heat is required to convert 5.0 g of water into steam.
Heat required for vaporization = mass of steam * heat of vaporization
= 5.0 g * 539 cal/g
Since energy is conserved, the heat lost by the copper object is equal to the heat gained by the water and calorimeter plus the heat required for vaporization.
Heat lost by copper object = Heat gained by water and calorimeter + Heat required for vaporization
Now, we can rearrange the equation to solve for the original temperature of the copper object.
Original temperature of copper object = (Heat lost by copper object - Heat gained by water and calorimeter) / (mass of copper object * specific heat capacity of copper)
Time to crunch the numbers and solve this mystery! I'll let you do the calculations and reveal the answer. Remember, every detective needs a little math to solve the case!
To find the original temperature of the copper object, we need to use the principle of energy conservation.
First, let's calculate the heat gained by the water and the calorimeter. We can use the formula:
Q = m * c * ΔT
Where:
Q is the heat gained/lost
m is the mass of the substance
c is the specific heat capacity of the substance
ΔT is the change in temperature
For the water and the calorimeter, the heat gained can be calculated as:
Qwater = (mass of water) * (specific heat capacity of water) * (change in temperature)
Qcalorimeter = (mass of calorimeter) * (specific heat capacity of copper) * (change in temperature)
The specific heat capacity of water is approximately 1 cal/g°C, and the specific heat capacity of copper is approximately 0.092 cal/g°C.
Given:
Mass of water (mwater) = 220 g
Mass of calorimeter (mcalorimeter) = 150 g
Change in temperature (ΔT) = Final temperature - Initial temperature = 100°C (boiling water)
Specific heat capacity of water (cwater) = 1 cal/g°C
Specific heat capacity of copper (ccopper) = 0.092 cal/g°C
Calculating the heat gained by the water and the calorimeter:
Qwater = (220 g) * (1 cal/g°C) * (100°C - 20°C)
Qcalorimeter = (150 g) * (0.092 cal/g°C) * (100°C - 20°C)
Now let's calculate the heat required to vaporize the water:
Qvaporization = (mass of steam) * (heat of vaporization)
Given:
Mass of steam (msteam) = 5 g
Heat of vaporization (Hvaporization) = 539 cal/g
Calculating the heat required to vaporize the water:
Qvaporization = (5 g) * (539 cal/g)
Now, according to the principle of energy conservation, the heat gained by the water and the calorimeter plus the heat required to vaporize the water should equal the heat lost by the copper object:
Qwater + Qcalorimeter + Qvaporization = -(mass of copper) * (specific heat capacity of copper) * (change in temperature)
We know:
Mass of copper (mcopper) = 300 g
Change in temperature (ΔT) = Final temperature - Initial temperature
Rearranging the equation and solving for the initial temperature of the copper object:
Initial temperature of the copper object = (Qwater + Qcalorimeter + Qvaporization) / (mcopper * ccopper) + Final temperature
Substituting the known values and solving the equation will give us the original temperature of the copper object.