A solid iron cylinder of mass 0.2 kg is quickly transferred from inside boiling water in a beaker at 100 C into 90g of water at 15 C contained in a calorimeter of mass 150g and specific heat capacity equal to 800 J/kg/K. Find the final temperature of the water.
(The Specific Heat Capacity of iron is 400 J/kg/K and that of water is 4200 J/kg/K)
The answer has to be 27 C. Can someone help me please?
The sum of the heats gained is zero (some will be negative heat gained, or absorbed).
Heat gained by iron+heat gained by calorimeter+heat gained by water=0
miron*Ciron(Tf-100)+mcalor*ccalor*(Tf-15)+masswater*cwater*(Tf-15)=0
do the algebra, solvr for Tf.
Sure, I can help you with that! To find the final temperature of the water, we can use the principle of conservation of energy.
The heat gained by the water is equal to the heat lost by the iron cylinder and the calorimeter. We can express this mathematically as:
Q_lost = Q_gained
The heat lost by the iron cylinder can be calculated using the formula:
Q_lost = mass_iron * specific_heat_iron * (initial_temperature_iron - final_temperature_iron)
Given that the mass of the iron cylinder is 0.2 kg, the specific heat capacity of iron is 400 J/kg/K, and the initial temperature of the iron was 100°C, we can calculate the heat lost by the iron cylinder.
Q_lost = 0.2 kg * 400 J/kg/K * (100°C - final_temperature_iron)
The heat gained by the water can be calculated using the formula:
Q_gained = mass_water * specific_heat_water * (final_temperature_water - initial_temperature_water)
Given that the mass of the water is 90 g (which is equivalent to 0.09 kg), the specific heat capacity of water is 4200 J/kg/K, and the initial temperature of the water is 15°C, we can calculate the heat gained by the water.
Q_gained = 0.09 kg * 4200 J/kg/K * (final_temperature_water - 15°C)
The heat lost by the calorimeter can be calculated using the formula:
Q_lost_calorimeter = mass_calorimeter * specific_heat_calorimeter * (initial_temperature_calorimeter - final_temperature_water)
Given that the mass of the calorimeter is 150 g (which is equivalent to 0.15 kg), the specific heat capacity of the calorimeter is 800 J/kg/K, and the initial temperature of the calorimeter is the same as the initial temperature of the water (15°C), we can calculate the heat lost by the calorimeter.
Q_lost_calorimeter = 0.15 kg * 800 J/kg/K * (15°C - final_temperature_water)
Now, we can set up the equation using the principle of conservation of energy:
Q_lost + Q_lost_calorimeter = Q_gained
0.2 kg * 400 J/kg/K * (100°C - final_temperature_iron) + 0.15 kg * 800 J/kg/K * (15°C - final_temperature_water) = 0.09 kg * 4200 J/kg/K * (final_temperature_water - 15°C)
Simplifying the equation:
80 (100 - final_temperature_iron) + 120 (15 - final_temperature_water) = 378 (final_temperature_water - 15)
Solving this equation will give us the final temperature of the water.
Once you substitute the given values into the equation and solve it, you should find that the final temperature of the water is 27°C.