A piece of copper ball of mass 30grams at 150 degree celcius is placed in a copper calorimeter of mass 70gams containing 60grams of waterof copper 40degree celcius if the final ofthe mixture is 50degree celcius calculate the specific heat capacity of copper (specific heat capacity of water=4ā¢2 joules per grams per kelvin
To calculate the specific heat capacity of copper, we can use the principle of energy conservation. The energy lost by the copper ball must be gained by the water and calorimeter.
First, let's calculate the energy lost by the copper ball using the equation:
šššš š” = ššš¢(šš ā šš)
Where:
šššš š” is the heat lost by the copper ball
š is the mass of the copper ball (30 grams)
šš¢ is the specific heat capacity of copper (unknown)
šš is the final temperature of the mixture (50 degrees Celsius)
šš is the initial temperature of the copper ball (150 degrees Celsius)
Substituting the given values into the equation:
šššš š” = 30g Ć šš¢ Ć (50Ā°C - 150Ā°C)
šššš š” = -3000šš¢
Next, let's calculate the energy gained by the water and calorimeter using the equation:
ššššššš = (šš¤ + šš) Ć šš¤(šš ā šš)
Where:
ššššššš is the heat gained by the water and calorimeter
šš¤ is the mass of water (60 grams)
šš is the mass of the calorimeter (70 grams)
šš¤ is the specific heat capacity of water (4.2 joules per gram per Kelvin)
šš is the final temperature of the mixture (50 degrees Celsius)
šš is the initial temperature of the water (40 degrees Celsius)
Substituting the given values into the equation:
ššššššš = (60g + 70g) Ć 4.2 J/gĀ°C Ć (50Ā°C - 40Ā°C)
ššššššš = 1320 J
Since energy is conserved, the energy lost by the copper ball equals the energy gained by the water and calorimeter:
-3000šš¢ = 1320
Now we can solve for the specific heat capacity of copper:
šš¢ = 1320 J / -3000
šš¢ ā -0.44 J/gĀ°C
The specific heat capacity of copper is approximately -0.44 joules per gram per Kelvin. Note that the negative sign indicates that copper releases heat when its temperature decreases.