A 500 g piece of silver at 150 degree Celsius is submerged in 1000 g of water at 5 degree Celsius to be cooled. determine the final temperature of silver and water given
Cwater = 4.18 x 10 ^3 J/kg degree Celsius
Csilver = 2.4 x 10^2 J/kg degree Celsius
HATING IT IS RIGHT STUPIDS
Well, imagine you have a magic potion that can instantly turn water into ice and silver into popsicles. Now, let's calculate the final temperature using this magical scenario!
First, let's determine the heat lost by the silver and gained by the water. We can use the equation:
Qlost = Qgained
The heat lost by the silver can be calculated using the formula:
Qlost = m x Csilver x ΔT
where m is the mass of the silver (0.5 kg), Csilver is the specific heat capacity of silver (2.4 x 10^2 J/kg degree Celsius), and ΔT is the change in temperature of the silver.
Similarly, the heat gained by the water can be calculated using the formula:
Qgained = m x Cwater x ΔT
where m is the mass of the water (1 kg), Cwater is the specific heat capacity of water (4.18 x 10^3 J/kg degree Celsius), and ΔT is the change in temperature of the water.
Since the silver is cooling down and the water is heating up, the temperature change for the silver will be negative, and for the water, it will be positive.
Let's assume the final temperature of both silver and water is T.
Now, let's substitute the known values into the equation:
0.5 x 2.4 x 10^2 x (T - 150) = 1 x 4.18 x 10^3 x (T - 5)
Simplifying the equation, we get:
1.2 x (T - 150) = 4.18 x (T - 5)
Now, let's solve for T:
1.2T - 180 = 4.18T - 20.9
2.98T = 159.1
T ≈ 53.4 degrees Celsius
So, my friend, according to these calculations, the final temperature of both the silver and water would be approximately 53.4 degrees Celsius. Just remember to keep that magic potion nearby in case you need to turn your silver into popsicles!
To find the final temperature of the silver and water mixture after they reach thermal equilibrium, we can use the principle of conservation of energy.
Let's assume that the final temperature of the mixture is 'T'.
The heat lost by the silver is equal to the heat gained by the water:
m1 * C1 * (Tf - Ti) = m2 * C2 * (Tf - Ti)
Where:
m1 = mass of silver = 500 g = 0.5 kg
C1 = specific heat capacity of silver = 2.4 x 10^2 J/kg degrees Celsius
Ti = initial temperature of the silver = 150 degrees Celsius
m2 = mass of water = 1000 g = 1 kg
C2 = specific heat capacity of water = 4.18 x 10^3 J/kg degrees Celsius
Ti = initial temperature of the water = 5 degrees Celsius
Now, let's solve the equation:
(0.5 kg) * (2.4 x 10^2 J/kg degrees Celsius) * (T - 150 degrees Celsius) = (1 kg) * (4.18 x 10^3 J/kg degrees Celsius) * (T - 5 degrees Celsius)
Simplifying the equation:
(0.5) * (2.4 x 10^2) * (T - 150) = (1) * (4.18 x 10^3) * (T - 5)
(1.2 x 10^2) * (T - 150) = (4.18 x 10^3) * (T - 5)
120T - 1.8 x 10^4 = 4.18 x 10^3T - 20.9 x 10^3
-18.9 x 10^3 = 4.06 x 10^3T - 120T
-18.9 x 10^3 = (4.06 x 10^3 - 120)T
T = (-18.9 x 10^3) / (4.06 x 10^3 - 120)
T ≈ 8.195 degrees Celsius
Therefore, the final temperature of the silver and water mixture is approximately 8.195 degrees Celsius.
To determine the final temperature of both the silver and water, we can use the principle of conservation of energy. The heat lost by the silver will be gained by the water, assuming no heat is lost to the surroundings.
The heat lost by the silver can be calculated using the formula:
Qsilver = m * Csilver * ΔTsilver
Where:
Qsilver = heat lost by silver (in Joules)
m = mass of silver (in kg)
Csilver = specific heat capacity of silver (in J/kg degree Celsius)
ΔTsilver = change in temperature of silver (final temperature - initial temperature)
Given values:
m = 0.5 kg
Csilver = 2.4 x 10^2 J/kg degree Celsius
ΔTsilver = final temperature of silver - initial temperature of silver
The heat gained by the water can be calculated using the formula:
Qwater = m * Cwater * ΔTwater
Where:
Qwater = heat gained by water (in Joules)
m = mass of water (in kg)
Cwater = specific heat capacity of water (in J/kg degree Celsius)
ΔTwater = change in temperature of water (final temperature - initial temperature)
Given values:
m = 1 kg (1000 g = 1 kg)
Cwater = 4.18 x 10^3 J/kg degree Celsius
ΔTwater = final temperature of water - initial temperature of water
Since the heat lost by the silver is equal to the heat gained by the water, we can equate the two equations:
Qsilver = Qwater
m * Csilver * ΔTsilver = m * Cwater * ΔTwater
Now, substituting the given values:
0.5 * 2.4 x 10^2 * ΔTsilver = 1 * 4.18 x 10^3 * ΔTwater
Simplifying the equation:
1.2 x 10^2 * ΔTsilver = 4.18 x 10^3 * ΔTwater
Now, we can solve for the final temperature. Rearranging the equation, we get:
ΔTsilver = (4.18 x 10^3 * ΔTwater) / (1.2 x 10^2)
Substituting the values:
ΔTsilver = (4.18 x 10^3 * ΔTwater) / 1.2 x 10^2
ΔTsilver = 34.83 * ΔTwater
Next, let's consider the change in temperature of the water. The final temperature of the water can be calculated using:
ΔTwater = (Tfinal - 5)
Substituting this value into the previous equation, we can solve for ΔTsilver:
ΔTsilver = 34.83 * (Tfinal - 5)
Finally, since we want to determine the final temperature of both the silver and the water, we set ΔTsilver equal to ΔTwater:
34.83 * (Tfinal - 5) = Tfinal - 150
Simplifying the equation:
34.83 * Tfinal - 174.15 = Tfinal - 150
34.83 * Tfinal - Tfinal = 174.15 - 150
33.83 * Tfinal = 24.15
Tfinal = 24.15 / 33.83
Tfinal ≈ 0.714 degree Celsius
Therefore, the final temperature of both the silver and water is approximately 0.714 degree Celsius.
Heat released by 0.5 kg silver=heat gained by 1 kg water upto final temp T.
Or 0.5(2.4x100)(150-T)=1(4.18x1000)(T-5)
Or 1.2(150-T)=41.8(T-5)
Or 180-1.2T=41.8T-209
Or 43T=389
Or T=9.05 deg Celsius