A metal ball of mass 1.8 kg with initial temperature of 46.5 C is dropped into a container of 3 kg of water at 15 C. If the final temperature is 22.6 C, what is the specific heat of the metal?
metal = ____ J/kg*C?
Well, it seems like that metal ball really wanted to take a dip in the water! Now, let's get to the bottom of this temperature mystery.
First, let's calculate the heat lost by the metal ball using the formula:
Q = m * c * ΔT
Where:
Q is the heat lost or gained
m is the mass of the metal ball
c is the specific heat of the metal
ΔT is the change in temperature
We know that the final temperature of the metal-water mix is 22.6°C and the initial temperature of the metal was 46.5°C. So, the ΔT value is:
ΔT = (22.6°C - 46.5°C) = -23.9°C
Taking into account the mass of the metal ball (m = 1.8 kg), we can calculate the heat lost:
Q = (1.8 kg) * c * ( -23.9°C)
Now, we also need to consider the heat gained by the water. Given that there are 3 kg of water in the container and the temperature change is from 15°C to 22.6°C, we can calculate the heat gained by the water as:
Q = (3 kg) * (specific heat of water) * (ΔT)
Since we're looking for the specific heat of the metal, we can equate these two equations:
(1.8 kg) * c * ( -23.9°C) = (3 kg) * (specific heat of water) * (22.6°C - 15°C)
Simplifying it further, we get:
c = [(3 kg) * (specific heat of water) * (22.6°C - 15°C)] / [(1.8 kg) * (-23.9°C)]
Now, my dear human, you just have to plug in the specific heat of water, do a few calculations, and you'll have your answer!
To find the specific heat of the metal, we can use the principle of energy conservation. The heat lost by the metal ball is equal to the heat gained by the water.
The heat lost by the metal ball can be calculated using the formula:
Q = mcΔT
Where:
Q = heat lost
m = mass of the metal ball
c = specific heat of the metal
ΔT = change in temperature
The heat gained by the water can be calculated using the formula:
Q = mcΔT
Where:
Q = heat gained
m = mass of the water
c = specific heat of water
ΔT = change in temperature
Since the heat lost by the metal ball is equal to the heat gained by the water, we can set the two equations equal to each other:
mcΔT = mcΔT
Substituting the given values:
(metal mass) * c * (final temperature - initial temperature) = (water mass) * c * (final temperature - water initial temperature)
We can now solve for the specific heat of the metal (c):
1.8 * c * (22.6 - 46.5) = 3 * c * (22.6 - 15)
Let's solve this equation step-by-step:
1.8 * c * (-23.9) = 3 * c * 7.6
-43.02c = 22.8c
Now, let's isolate c:
-43.02c - 22.8c = 0
-65.82c = 0
c = 0 / -65.82
c = 0
Therefore, the specific heat of the metal is 0 J/kg°C.
To find the specific heat of the metal, we can use the principle of heat transfer.
The heat gained by the metal can be calculated using the formula:
Q = m * c * ΔT
Where:
Q is the heat gained by the metal (in Joules),
m is the mass of the metal (in kg),
c is the specific heat capacity of the metal (in J/kg*C), and
ΔT is the change in temperature (final temperature - initial temperature).
To solve for the specific heat capacity, we rearrange the formula:
c = Q / (m * ΔT)
First, let's calculate the heat gained by the metal.
We have the initial temperature of the metal (46.5°C), the final temperature (22.6°C), and the mass (1.8 kg).
ΔT = 22.6°C - 46.5°C = -23.9°C
Next, we calculate the heat gained by the metal:
Q = m * c * ΔT
Since the specific heat capacity of the water is generally much larger than that of the metal, we can neglect the heat transferred to the water by the metal.
Q = 0 (heat transferred to the water)
Therefore, the heat gained by the metal is zero.
Now, we can substitute the values into the formula to find the specific heat capacity of the metal:
c = Q / (m * ΔT)
c = 0 / (1.8 kg * -23.9°C)
c = 0
Based on the calculations, the specific heat of the metal is 0 J/kg*C. However, this may not be a physical or realistic value. It is possible that the specific heat capacity of the metal was not provided or the calculations were done incorrectly.