The soup in a bowl is too hot to eat. Although there are no ice cubes in the freezer, there are several stainless steel spoons that have been stored in the freezer for several hours. By placing the cold spoons in the hot soup, the soup's temperature is lowered from a temperature of 82�Ž to 48�Ž. The mass of the soup is 0.1kg while the mass of each spoon is 0.04kg.
a) Assuming that soup has a specific heat capacity of 4186J/(kg�Ž),how much heat is transferred from the soup?
b) If the initial temperature of the spoons is -15�Žand their specific heat capacity is 448J/(kg�Ž), how many spoons are needed to cool the soup? (Hint: Their final temp=48�Ž)
14232.4J
c zxv
To solve these questions, we need to use the equation Q = mcΔT, where Q represents the heat transferred, m represents the mass, c represents the specific heat capacity, and ΔT represents the change in temperature.
a) To find the heat transferred from the soup, we can use the equation Q = mcΔT.
Given:
Specific heat capacity of the soup (c) = 4186 J/(kg�Ž)
Mass of the soup (m) = 0.1 kg
Initial temperature of the soup = 82�Ž
Final temperature of the soup = 48�Ž
Using the equation Q = mcΔT, we can calculate the heat transferred:
Q = (0.1 kg) * (4186 J/(kg�Ž)) * (48�Ž - 82�Ž)
First, let's find the difference in temperature:
ΔT = (48�Ž - 82�Ž) = -34�Ž
Now we can calculate the heat transferred:
Q = (0.1 kg) * (4186 J/(kg�Ž)) * (-34�Ž)
Q = -14272.4 J
Therefore, the heat transferred from the soup is -14272.4 J (rounded to 1 decimal place). The negative sign indicates that heat is being lost by the soup.
b) To find the number of spoons needed to cool the soup, we can use the same equation Q = mcΔT.
Given:
Specific heat capacity of the spoons (c) = 448 J/(kg�Ž)
Initial temperature of the spoons = -15�Ž
Final temperature of the spoons = 48�Ž
Using the equation Q = mcΔT, we can calculate the heat transferred from the spoons:
Q = (n * 0.04 kg) * (448 J/(kg�Ž)) * (48�Ž - (-15�Ž))
First, let's find the difference in temperature:
ΔT = (48�Ž - (-15�Ž)) = 63�Ž
Now, let's substitute the values into the equation and solve for "n" (the number of spoons needed):
Q = (n * 0.04 kg) * (448 J/(kg�Ž)) * 63�Ž
Q = An unknown value
From part a, we know that the heat transferred is -14272.4 J. So, we can set up the equation:
-14272.4 J = (n * 0.04 kg) * (448 J/(kg�Ž)) * 63�Ž
Divide both sides of the equation by (0.04 kg * 448 J/(kg�Ž) * 63�Ž) to solve for "n":
-14272.4 J / (0.04 kg * 448 J/(kg�Ž) * 63�Ž) = n
Calculating the right side of the equation:
n ≈ 3.01 spoons
Therefore, approximately 3 spoons (rounded to the nearest whole number) are needed to cool the soup from 82�Ž to 48�Ž.