By how much will the temperature of 2.00kg of air change if it absorbs all of the heat given up when 1.00kg of water cools from 40.degree to 20 degree?
Thanks for all your help
Why did the air go on a diet? Because it wanted to be cooler!
To calculate the change in temperature of the air, we can use the concept of specific heat capacity. The specific heat capacity of water is 4.18 J/g°C, and the specific heat capacity of air is roughly 1 J/g°C.
First, let's find out how much heat is given up by the water when it cools from 40°C to 20°C:
Heat lost by water = mass of water × specific heat capacity of water × change in temperature
Heat lost by water = 1.00 kg × 4.18 J/g°C × (40 - 20)°C
Now, let's calculate the change in temperature of the air:
Change in temperature of air = Heat lost by water / (mass of air × specific heat capacity of air)
Change in temperature of air = (1.00 kg × 4.18 J/g°C × (40 - 20)°C) / (2.00 kg × 1 J/g°C)
Calculating that out, the change in temperature of the air turns out to be approximately 10°C. So, the temperature of the air would decrease by about 10 degrees.
Just remember, if you're ever feeling hot, just hang out with some water, and the air will cool down around you!
To solve this problem, we need to use the specific heat capacity formula:
Q = mcΔT
Where:
Q = heat energy transferred (in joules)
m = mass (in kilograms)
c = specific heat capacity (in joules per kilogram-degree Celsius)
ΔT = change in temperature (in degrees Celsius)
Given:
Mass of air (m₁) = 2.00 kg
Mass of water (m₂) = 1.00 kg
Initial temperature of water (T₂i) = 40.0 °C
Final temperature of water (T₂f) = 20.0 °C
First, we need to calculate the energy transferred when the water cools using the formula:
Q = m₂c₂ΔT₂
We know that the specific heat capacity of water is approximately 4,184 J/(kg°C).
Using the formula, we can calculate the energy transferred:
Q = (1.00 kg)(4,184 J/(kg°C))(20.0 °C - 40.0 °C)
Simplifying the equation:
Q = (1.00 kg)(4,184 J/(kg°C))(-20.0 °C)
Q = -83,680 J
Now we can use the energy transferred by the water (Q) to find the change in temperature (ΔT₁) of the air.
Q = m₁c₁ΔT₁
Rearranging the formula to solve for ΔT₁:
ΔT₁ = Q/(m₁c₁)
The specific heat capacity of air (c₁) is approximately 1,005 J/(kg°C).
Substituting the values:
ΔT₁ = -83,680 J / (2.00 kg)(1,005 J/(kg°C))
= -83,680 J / 2,010 J/°C
≈ -41.6 °C
Finally, the change in temperature of the 2.00 kg of air would be approximately -41.6 °C if it absorbs all the heat given up when 1.00 kg of water cools from 40.0 °C to 20.0 °C.
To determine the change in temperature of the air when it absorbs the heat given up by the water, we can use the concept of specific heat capacity.
Specific heat capacity is defined as the amount of heat energy required to raise the temperature of a given mass of substance by 1 degree Celsius (or 1 Kelvin).
Given:
Mass of water (m1) = 1.00 kg
Change in temperature of water (ΔT1) = 40 degree - 20 degree = 20 degree
We assume that the specific heat capacity of water (c1) is approximately 4.18 kJ/kg°C (this value may vary depending on the state of the water).
We need to calculate the heat energy (Q1) released by the water using the following formula:
Q1 = (m1) x (c1) x (ΔT1)
Q1 = 1.00 kg x 4.18 kJ/kg°C x 20°C
Q1 = 83.6 kJ
Now, we need to determine how much the temperature of the air will change. For that, we need to know the specific heat capacity of air (c2). The specific heat capacity of air is approximately 1.00 kJ/kg°C.
Since the air absorbs all of the heat energy released by the water, the heat absorbed by the air is equal to the heat released by the water. Therefore, we have:
Q2 = Q1 = 83.6 kJ
Now, we can calculate the change in temperature of the air (ΔT2) using the following formula:
ΔT2 = Q2 / [(m2) x (c2)]
Given:
Mass of air (m2) = 2.00 kg
Specific heat capacity of air (c2) = 1.00 kJ/kg°C
ΔT2 = 83.6 kJ / (2.00 kg x 1.00 kJ/kg°C)
ΔT2 = 41.8°C
Therefore, the temperature of the 2.00 kg of air will increase by 41.8 degrees Celsius when it absorbs all of the heat given up when 1.00 kg of water cools from 40 degrees to 20 degrees.
q = 1,000g H2O x specific heat H2O x delta T
Solve for q and substitute for air as
q = 2000g air x specific heat air x delta T.
Solve for delta T.