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!
To determine the final temperature of the object, we can use the principle of conservation of energy.
First, let's calculate the heat gained or lost by the object in the initial step when it is placed in the 40 grams of water at 100 degrees Celsius. We can use the formula:
Q1 = mcΔT1
where Q1 is the heat gained or lost by the object, m is the mass of the water, c is the specific heat capacity of water, and ΔT1 is the change in temperature.
Given:
m = 40 g
c (specific heat capacity of water) = 1 calorie/gram Celsius
ΔT1 = 100 - 20 = 80°C
Q1 = (40 g) * (1 cal/g°C) * (80°C) = 3200 calories
The object loses 3200 calories of heat energy when placed in the hot water.
Next, let's calculate the heat gained or lost by the object in the second step when it is placed in the 200 grams of water at 0 degrees Celsius. We can use the same formula:
Q2 = mcΔT2
where Q2 is the heat gained or lost by the object, m is the mass of the water, c is the specific heat capacity of water, and ΔT2 is the change in temperature.
Given:
m = 200 g
c (specific heat capacity of water) = 1 calorie/gram Celsius
ΔT2 = Tf - 0, where Tf is the final temperature of the object
Q2 = (200 g) * (1 cal/g°C) * (Tf - 0) = 200 Tf calories
Now, according to the principle of conservation of energy, the heat lost by the object in the first step (Q1) is equal to the heat gained by the object in the second step (Q2).
Therefore, Q1 = Q2
3200 calories = 200 Tf calories
Solving for Tf:
Tf = (3200 calories) / (200 g) = 16°C
Hence, the final temperature of the object is 16 degrees Celsius when it is placed in the 200 grams of water at 0 degrees Celsius.
Note: The heat capacity of the object (100 calories/Celsius) was given but not utilized in the calculation, as we assumed that the heat gained or lost by the object is negligible compared to the heat gained or lost by the water.