A piece of copper weighing 400g is heated to 100°c and then quickly transferred to a copper calorimeter of Mass of 10g containing 100g of liquid of unknown specific heat capacity at 30°c if the final temperature of the mixture is 50°c calculate the specific heat capacity of liquid ( specific heat capacity if copper = 390j/kg/k)
[mass Cu sample x specific heat Cu x (Tfinal-Tinitial)] + [mass liquid + specific heat liquid x (Tfinal-Tinitial)] + [mass container of Cu x specific heat Cu x (Tfinal-Tinitial)] = 0
Plus in the numbers and solve for specific heat liquid.
Temperature can be confusing so here is a summary of the data.
400 g Cu Tfinal-Tinitial. Start at 100 C and ends at 50 C.
10 g container of Cu. Tfinal-Tinitial. Start at 30C and ends at 50 C.
100 g liquid. Tfinal - Tinitial. Starts at 30 C and ends at 50 C.
To solve this problem, we can use the principle of conservation of energy. The heat gained by the liquid and the calorimeter is equal to the heat lost by the copper.
Step 1: Calculate the heat gained by the liquid and the calorimeter.
The heat gained by the liquid and the calorimeter can be calculated using the formula Q = mcΔT, where Q is the heat gained, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
For the liquid and the calorimeter:
Q1 = (mass of liquid + mass of calorimeter) x specific heat capacity of the liquid x ΔT1
Q1 = (100g + 10g) x c x (50°C - 30°C)
Step 2: Calculate the heat lost by the copper.
The heat lost by the copper can be calculated using the formula Q = mcΔT, where Q is the heat lost, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
Q2 = mass of copper x specific heat capacity of copper x ΔT2
Q2 = 400g x 390 J/kg/°C x (100°C - 50°C)
Step 3: Set the heat gained equal to the heat lost.
Q1 = Q2
(100g + 10g) x c x (50°C - 30°C) = 400g x 390 J/kg/°C x (100°C - 50°C)
Simplifying the equation:
(110g) x c x 20°C = 400g x 390 J/kg/°C x 50°C
Multiplying the values:
2200c = 7800000
Dividing both sides by 2200:
c = 7800000 / 2200
c ≈ 3545.45 J/kg/°C
Therefore, the specific heat capacity of the liquid is approximately 3545.45 J/kg/°C.
To calculate the specific heat capacity of the liquid, we can use the principle of conservation of energy. The heat gained by the liquid will be equal to the heat lost by the copper.
The equation for heat energy is given by:
Q = mcΔT
Where:
Q represents the heat energy (in joules),
m represents the mass (in kilograms),
c represents the specific heat capacity (in J/kg/°C),
and ΔT represents the change in temperature (in °C).
First, let's calculate the heat lost by the copper:
Q1 = mcΔT
= (0.4 kg) * (390 J/kg/°C) * (100°C - 50°C)
= 0.4 kg * 390 J/kg/°C * 50°C
= 7,800 J
Next, let's calculate the heat gained by the liquid:
Q2 = mcΔT
= (0.1 kg + 0.1 kg) * (c) * (50°C - 30°C)
= 0.2 kg * c * 20°C
= 4c J
Since the heat lost by the copper is equal to the heat gained by the liquid, we can set up an equation:
7,800 J = 4c J
Now we can solve for c:
c = 7,800 J / 4 J
c = 1,950 J/kg/°C
Therefore, the specific heat capacity of the liquid is 1,950 J/kg/°C.