Please Help!
An object at 20 degrees Celsius with heat capacity 100 calories/Celsius is placed in 40 grams of water at 100 degrees Celsius. After equilibrium is reached, the object is removed from the hot water and placed in 200 grams of water at 0 degrees Celsius. What is the final temperature of the object? Explain.
Thanks for the help!
Why did the object go on a hot water vacation? It wanted to chill out and cool down in a new environment!
But don't worry, I'm here to help you with your question.
To solve this problem, we can use the principle of energy conservation - the heat lost by the hot water should be equal to the heat gained by the object.
First, let's calculate the amount of heat lost by the hot water. We can use the equation 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. The mass of the water is 40 grams, and the specific heat capacity of water is approximately 1 calorie/gram/degree Celsius. The change in temperature is 100 degrees Celsius (initial temperature of 100 minus final temperature). So, the heat lost by the hot water is 40 * 1 * 100 calories.
Now, let's calculate the heat gained by the object. The heat capacity of the object is given as 100 calories/degree Celsius. Since the object is initially at 20 degrees Celsius and ends up at the final temperature, we can calculate the heat gained as 100 * (final temperature - 20) calories.
According to the principle of energy conservation, the heat lost by the hot water should be equal to the heat gained by the object. Therefore, we can equate the two equations above:
40 * 1 * 100 = 100 * (final temperature - 20)
Now, solve for the final temperature:
4000 = 100 * (final temperature - 20)
Divide both sides by 100:
40 = final temperature - 20
Add 20 to both sides:
final temperature = 40 + 20 = 60 degrees Celsius
Therefore, the final temperature of the object is 60 degrees Celsius. It's not too hot or too cold, just the right amount of chill after its hot water adventure!
To find the final temperature of the object, we can use the principle of conservation of energy.
First, let's calculate the heat transferred from the object (at 20 degrees Celsius) to the hot water (at 100 degrees Celsius) until equilibrium is reached.
The formula to calculate the heat transfer is:
Q = mcΔT
where Q is the heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
Given:
mass of the object (m1) = unknown
specific heat capacity of the object (c1) = 100 calories/Celsius
initial temperature of the object (T1) = 20 degrees Celsius
mass of the hot water (m2) = 40 grams
specific heat capacity of water (c2) = 1 calorie/gram Celsius
initial temperature of the hot water (T2) = 100 degrees Celsius
We need to find ΔT1, which is the change in temperature for the object.
Using the heat transfer formula:
Q = mcΔT
Q1 = m1c1ΔT1
The heat transfer for the object is equal to the heat absorbed by the water, so:
Q1 = Q2
m1c1ΔT1 = m2c2ΔT2
Substituting the given values:
m1 * 100 * ΔT1 = 40 * 1 * (100 - T1)
100m1ΔT1 = 40(100 - T1)
Simplifying the equation:
100m1ΔT1 = 4000 - 40T1
100m1ΔT1 + 40T1 = 4000
ΔT1 = 40 (40 - m1)
Now, the object is removed from the hot water and placed in 200 grams of water at 0 degrees Celsius. The heat transfer equation is:
Q3 = mcΔT
Q3 = m1c1ΔT1 = m3c3ΔT3
We know:
m1c1ΔT1 = Q3
m3c3ΔT3 = Q3
Since Q3 is the same for both sides of the equation:
m1c1ΔT1 = m3c3ΔT3
Substituting the given values:
m1 * 100 * ΔT1 = 200 * 1 * (T3 - 0)
m1 * 100 * (40 - m1) = 200T3
Simplifying the equation:
4000m1 - 100m1^2 = 200T3
100m1^2 - 4000m1 + 200T3 = 0
This is a quadratic equation in terms of m1, the mass of the object. Solving for m1 using the quadratic formula:
m1 = (-b ± √(b^2 - 4ac)) / 2a
Using a = 100, b = -4000, and c = 200T3:
m1= (-(-4000) ± √((-4000)^2 - 4*100*200T3)) / 2(100)
m1 = (4000 ± √(16000000 - 80000T3)) / 200
We can simplify the equation further to:
m1 = (20 ± √(400 - 2T3)) / 1
Since mass cannot be negative, we take the positive root:
m1 = (20 + √(400 - 2T3))
We know that the mass of the object cannot be more than the total mass of water before equilibrium, which is 40g. Therefore, we can set up the inequality:
m1 ≤ m2
(20 + √(400 - 2T3)) ≤ 40
Solving for T3, the final temperature of the object:
20 + √(400 - 2T3) ≤ 40
√(400 - 2T3) ≤ 20
400 - 2T3 ≤ 400
-2T3 ≤ 0
T3 ≥ 0
Since 0 degrees Celsius is the lower limit for the final temperature of the object, the final temperature of the object is 0 degrees Celsius.
Therefore, after being removed from the hot water at equilibrium, the object cools down to the same temperature as the cold water (0 degrees Celsius).
To find the final temperature of the object, we need to use the principle of conservation of energy. The heat gained by the cold water will be equal to the heat lost by the object.
First, let's calculate the heat gained by the cold water. We can use the formula:
Heat gained = mass of water * specific heat capacity of water * change in temperature
Given:
Mass of water = 200 grams
Specific heat capacity of water = 1 calorie/gram Celsius
Change in temperature = final temperature - initial temperature = final temperature - 0 (because the cold water is initially at 0 degrees Celsius)
Heat gained by the cold water = 200 * 1 * (final temperature - 0) calories
Now, let's calculate the heat lost by the object. We can use the formula:
Heat lost = heat capacity of the object * change in temperature
Given:
Heat capacity of the object = 100 calories/Celsius
Change in temperature = initial temperature - final temperature = 100 - final temperature
Heat lost by the object = 100 * (100 - final temperature) calories
Since the principle of conservation of energy states that the heat gained is equal to the heat lost, we can set up the following equation:
Heat gained by the cold water = Heat lost by the object
200 * 1 * (final temperature - 0) = 100 * (100 - final temperature)
Now, we solve this equation to find the final temperature of the object.
200 * (final temperature) = 10000 - 100 * (final temperature)
200 * (final temperature) + 100 * (final temperature) = 10000
300 * (final temperature) = 10000
(final temperature) = 10000 / 300
(final temperature) ≈ 33.33 degrees Celsius
So, the final temperature of the object is approximately 33.33 degrees Celsius.