A piece of copper in 400 gram is heated to hundred degree celsius and then quickly transferred to a copper calorimeter of mass 10 g containing 100 g of liquid of unknown specific heat capacity at 30 degree celsius if the final temperature of the mixture is 50 degree celsius
calculate
the specific heat capacity of the liquid given that the specific heat capacity of copper is 390 joules per kg per Kelvin
To calculate the specific heat capacity of the liquid, we can use the principle of heat transfer. The heat gained by the liquid is equal to the heat lost by the copper.
First, let's calculate the heat lost by the copper:
q1 = m1 * c1 * ΔT1
Where:
q1 = heat lost by the copper
m1 = mass of the copper (400 g)
c1 = specific heat capacity of copper (390 J/kgK)
ΔT1 = change in temperature of the copper (final temp - initial temp = 50°C - 100°C = -50 K)
q1 = 400 g * (1 kg / 1000 g) * 390 J/(kgK) * (-50 K)
q1 = -7800 J
Note: The negative sign indicates heat loss.
Next, let's calculate the heat gained by the liquid:
q2 = m2 * c2 * ΔT2
Where:
q2 = heat gained by the liquid
m2 = mass of the liquid (100 g)
c2 = specific heat capacity of the liquid (to be determined)
ΔT2 = change in temperature of the liquid (final temp - initial temp = 50°C - 30°C = 20 K)
q2 = 100 g * (1 kg / 1000 g) * c2 * 20 K
q2 = (0.02 kg) * c2 * 20 K
q2 = 0.4 c2 K
Now, since heat gained by the liquid is equal to the heat lost by the copper, we can equate q1 and q2:
-7800 J = 0.4 c2 K
To find the specific heat capacity of the liquid (c2), we can rearrange the equation:
c2 = -7800 J / (0.4 K)
c2 = -19500 J/(kgK)
The specific heat capacity of the liquid is -19500 J/(kgK).