A piece of copper weighing 400g is heated to 100 degree celcius and quickly transferred to a copper calorimeter of mass 10g containing 100g of liquid of unknown specific heat capacity at 30 degree celcius.if the final temperature of the mixture is 50 degree celcius,calculate the specific heat capacity of the liquid.(specific heat capacity of copper is 390J/KG/K)
To solve this problem, we can use the principle of conservation of energy.
The heat gained by the liquid can be calculated using the equation:
Heat gained by liquid = Heat lost by copper
The heat gained or lost in a substance can be calculated using the formula:
Q = mcΔT
Where:
Q is the heat gained or lost
m is the mass
c is the specific heat capacity
ΔT is the change in temperature
Now let's calculate the heat lost by the copper:
Q(copper) = mc(copper)ΔT
Given:
Mass of the copper = 400g = 0.4kg
Specific heat capacity of copper = 390J/kg/K
Initial temperature of the copper = 100°C
Final temperature of the mixture = 50°C
ΔT(copper) = Final temperature of the mixture - Initial temperature of the copper
= 50°C - 100°C
= -50°C
Since the copper lost heat, we take the absolute value of ΔT.
ΔT(copper) = |-50°C| = 50°C
Now let's calculate the heat gained by the liquid:
Q(liquid) = mc(liquid)ΔT
Given:
Mass of the liquid = 100g = 0.1kg
Specific heat capacity of the liquid = ?
ΔT(liquid) = Final temperature of the mixture - Initial temperature of the liquid
= 50°C - 30°C
= 20°C
Now, we can equate the heat gained by the liquid to the heat lost by the copper:
Q(liquid) = Q(copper)
mc(liquid)ΔT(liquid) = mc(copper)ΔT(copper)
0.1kg * c(liquid) * 20°C = 0.4kg * 390J/kg/K * 50°C
Now, we can solve for the specific heat capacity of the liquid (c(liquid)):
c(liquid) = (0.4kg * 390J/kg/K * 50°C) / (0.1kg * 20°C)
c(liquid) ≈ 780 J/kg/K
Therefore, the specific heat capacity of the liquid is approximately 780 J/kg/K.