100 grams of a liquid with a temp of 70 degreess celcius and a specific heat of 4 J/K/gram is poured into an aluminum pan having a mass of 4000 grams and an initial temp of 20 degrees. with a specific heat capacity of 0.9. what is the final temp of the liquid in the pan after it reaches equilibrium. assume no heat is lost to the surroudnings.
To find the final temperature of the liquid after reaching equilibrium, we can use the principle of conservation of energy.
The heat gained by the aluminum pan equals the heat lost by the liquid.
Heat gained by the aluminum pan:
Q1 = m1 * c1 * ΔT1
Where:
m1 = mass of the pan = 4000 grams
c1 = specific heat capacity of aluminum = 0.9 J/g°C
ΔT1 = change in temperature of the pan = Final temperature - Initial temperature = Final temperature - 20°C
Heat lost by the liquid:
Q2 = m2 * c2 * ΔT2
Where:
m2 = mass of the liquid = 100 grams
c2 = specific heat capacity of the liquid = 4 J/g°C
ΔT2 = change in temperature of the liquid = Final temperature - 70°C
Since no heat is lost to the surroundings, Q1 = Q2
Therefore, m1 * c1 * ΔT1 = m2 * c2 * ΔT2
(4000 grams) * (0.9 J/g°C) * (Final temperature - 20°C) = (100 grams) * (4 J/g°C) * (Final temperature - 70°C)
Simplifying the equation:
3600 * (Final temperature - 20°C) = 400 * (Final temperature - 70°C)
3600 * Final temperature - 3600 * 20°C = 400 * Final temperature - 400 * 70°C
Expanding and rearranging the equation:
3600 * Final temperature - 72000°C = 400 * Final temperature - 28000°C
3600 * Final temperature - 400 * Final temperature = 72000°C - 28000°C
3200 * Final temperature = 44000°C
Final temperature = 44000°C / 3200
Final temperature = 13.75°C
Therefore, the final temperature of the liquid in the pan after reaching equilibrium is approximately 13.75°C.
To find the final temperature of the liquid in the pan after reaching equilibrium, we can use the principle of conservation of energy. The heat lost by the liquid will be equal to the heat gained by the pan.
Let's start by calculating the heat lost by the liquid using the formula:
Q1 = m1 * c1 * ΔT1
Where:
Q1 = heat lost by the liquid
m1 = mass of the liquid (100 grams)
c1 = specific heat capacity of the liquid (4 J/K/gram)
ΔT1 = change in temperature of the liquid (final temperature - initial temperature)
Now, let's calculate the heat gained by the pan using the formula:
Q2 = m2 * c2 * ΔT2
Where:
Q2 = heat gained by the pan
m2 = mass of the pan (4000 grams)
c2 = specific heat capacity of the pan (0.9 J/g/°C)
ΔT2 = change in temperature of the pan (final temperature - initial temperature)
Since the heat lost by the liquid is equal to the heat gained by the pan, we have:
Q1 = Q2
m1 * c1 * ΔT1 = m2 * c2 * ΔT2
Now, let's solve for ΔT1 and ΔT2:
ΔT1 = (m2 * c2 * ΔT2) / (m1 * c1)
Substituting the given values into the equation:
ΔT1 = (4000 g * 0.9 J/g/°C * (20 °C - X)) / (100 g * 4 J/K/g)
Where X represents the final temperature of the liquid in the pan.
Now, let's solve for X:
ΔT1 = (3600 (20 - X)) / 400
3600 (20 - X) = 400 ΔT1
3600 (20 - X) = 400 ((20 - X) - 20)
3600 (20 - X) = 400 (20 - X - 20)
3600 (20 - X) = 400 (-X)
72,000 - 3600X = -400X
72,000 = 3200X
X = 72,000 / 3200
X = 22.5
Hence, the final temperature of the liquid in the pan after reaching equilibrium is 22.5 degrees Celsius.