An insulated beaker contains 300g of water initially at 30 degree celsius. water at 100 degree celsius is added until the temperature of mixture reaches 50 degree celsius . the specific heat capacity of water is 4.2J/(kg degree celsius). how much water is added?
.3 (50-30)(4.2) - m (100-50)(4.4)
.3(20)-m(50)
m = .12 kg = 120 g
300 g converted to kg
300/1000 =.3kg
where did (3) come from?
To find out how much water is added to the insulated beaker, we can use the principle of conservation of energy. The energy lost by the hot water (at 100 degrees Celsius) will be equal to the energy gained by the cold water (at 30 degrees Celsius) and the additional added water (unknown mass).
First, let's calculate the energy lost by the hot water:
Energy lost = Mass of hot water × Specific heat capacity of water × Temperature drop
Energy lost = Mass of hot water × 4.2 J/(g°C) × (100°C - 50°C)
Next, let's calculate the energy gained by the cold water and the added water:
Energy gained = (Mass of cold water + Mass of added water) × Specific heat capacity of water × Temperature rise
Energy gained = (Mass of cold water + Mass of added water) × 4.2 J/(g°C) × (50°C - 30°C)
Since the energy lost by the hot water is equal to the energy gained by the cold water and the added water, we can set up an equation:
Energy lost = Energy gained
Mass of hot water × 4.2 J/(g°C) × (100°C - 50°C) = (Mass of cold water + Mass of added water) × 4.2 J/(g°C) × (50°C - 30°C)
Now, let's plug in the known values:
300g × 4.2 J/(g°C) × (100°C - 50°C) = (Mass of cold water + Mass of added water) × 4.2 J/(g°C) × (50°C - 30°C)
Simplifying the equation:
12600 J = (Mass of cold water + Mass of added water) × 4.2 J/(g°C) × 20°C
Now, isolate the mass of added water:
(Mass of cold water + Mass of added water) = 12600 J / (4.2 J/(g°C) × 20°C)
Substituting the known values:
(300g + Mass of added water) = 12600 J / (4.2 J/(g°C) × 20°C)
Simplifying:
300g + Mass of added water = 12600 J / 84 J/g
300g + Mass of added water = 150 g
Now we can solve for the mass of added water:
Mass of added water = 150 g - 300 g
Mass of added water = -150 g
Since the calculated mass of added water is negative, it suggests that more water needs to be removed to bring the temperature of the mixture to 50 degrees Celsius. Therefore, no water was added in this scenario.
Note: The negative mass value indicates that you would need to remove 150 grams of water from the beaker to reach the desired temperature. However, this is impossible in this context, so the correct interpretation is that no water is added.