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!