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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!
How much energy is removed to cool 50.0 g of steam at 110 degrees Celsius to ice at -10 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 cooler water will be equal to the heat lost by the hotter object. We can use the formula:
Q = mcΔT
Where:
Q = heat gained or lost (in calories)
m = mass of the substance (in grams)
c = specific heat capacity (in calories/gram°C)
ΔT = change in temperature (in °C)
First, let's calculate the heat lost by the object when it is placed in 40 grams of water at 100 degrees Celsius.
Q_lost = m_object * c_object * ΔT_object
Where:
m_object = mass of the object (unknown)
c_object = specific heat capacity of the object (100 calories/°C)
ΔT_object = change in temperature of the object (unknown)
Next, let's calculate the heat gained by the cooler water when the object is placed in it at 0 degrees Celsius.
Q_gained = m_water * c_water * ΔT_water
Where:
m_water = mass of the water (200 grams)
c_water = specific heat capacity of water (1 calorie/°C)
ΔT_water = change in temperature of the water (unknown)
Since the heat lost by the object (Q_lost) is equal to the heat gained by the water (Q_gained), we can equate the two equations:
m_object * c_object * ΔT_object = m_water * c_water * ΔT_water
Substituting the given values:
(unknown) * 100 * ΔT_object = 200 * 1 * (final temperature of water - 0)
Simplifying the equation, we can divide both sides by 100:
ΔT_object = 2 * (final temperature of water - 0)
Therefore, the final temperature of the object will be twice the change in temperature of the water. To find the final temperature of the object, we need to determine the change in temperature of the water (ΔT_water).
Using the formula:
ΔT_water = final temperature of the water - initial temperature of the water
Given:
Initial temperature of the water = 0 degrees Celsius
Final temperature of the water = unknown
Since the object is at equilibrium with the water when it is removed, the final temperature of the water is the same as the initial temperature of the object, which is 20 degrees Celsius.
Substituting this value into the equation:
ΔT_water = final temperature of the water - initial temperature of the water
ΔT_water = final temperature of the water - 0
ΔT_water = final temperature of the water
Now we can substitute this equation back into the first equation:
ΔT_object = 2 * ΔT_water
ΔT_object = 2 * final temperature of the water
Since we know that ΔT_object = 20 - initial temperature of the object:
20 - initial temperature of the object = 2 * final temperature of the water
Solving for the final temperature of the water:
final temperature of the water = (20 - initial temperature of the object) / 2
Given:
Initial temperature of the object = 20 degrees Celsius
final temperature of the water = (20 - 20) / 2 = 0 / 2 = 0 degrees Celsius
Therefore, the final temperature of the object will also be 0 degrees Celsius.