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=4•2 joules per grams per kelvin

Question

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=4•2 joules per grams per kelvin

Answer 1

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:

𝑄𝑙𝑜𝑠𝑡 = 𝑚𝑐𝑢(𝑇𝑓 − 𝑇𝑖)

Where:
𝑄𝑙𝑜𝑠𝑡 is the heat lost by the copper ball
𝑚 is the mass of the copper ball (30 grams)
𝑐𝑢 is the specific heat capacity of copper (unknown)
𝑇𝑓 is the final temperature of the mixture (50 degrees Celsius)
𝑇𝑖 is the initial temperature of the copper ball (150 degrees Celsius)

Substituting the given values into the equation:

𝑄𝑙𝑜𝑠𝑡 = 30g × 𝑐𝑢 × (50°C - 150°C)
𝑄𝑙𝑜𝑠𝑡 = -3000𝑐𝑢

Next, let's calculate the energy gained by the water and calorimeter using the equation:

𝑄𝑔𝑎𝑖𝑛𝑒𝑑 = (𝑚𝑤 + 𝑚𝑐) × 𝑐𝑤(𝑇𝑓 − 𝑇𝑖)

Where:
𝑄𝑔𝑎𝑖𝑛𝑒𝑑 is the heat gained by the water and calorimeter
𝑚𝑤 is the mass of water (60 grams)
𝑚𝑐 is the mass of the calorimeter (70 grams)
𝑐𝑤 is the specific heat capacity of water (4.2 joules per gram per Kelvin)
𝑇𝑓 is the final temperature of the mixture (50 degrees Celsius)
𝑇𝑖 is the initial temperature of the water (40 degrees Celsius)

Substituting the given values into the equation:

𝑄𝑔𝑎𝑖𝑛𝑒𝑑 = (60g + 70g) × 4.2 J/g°C × (50°C - 40°C)
𝑄𝑔𝑎𝑖𝑛𝑒𝑑 = 1320 J

Since energy is conserved, the energy lost by the copper ball equals the energy gained by the water and calorimeter:

-3000𝑐𝑢 = 1320

Now we can solve for the specific heat capacity of copper:

𝑐𝑢 = 1320 J / -3000
𝑐𝑢 ≈ -0.44 J/g°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.