A substance has the following properties; Boiling Point = 140 °C, Freezing Point = -10 °C, specific heat as a gas, liquid and solid, 0.3 cal/g.C°, 1.8 cal/g.C° and 0.8 cal/g.C° respectively. Heat of fusion =100 cal/g and heat of vaporization = 480 cal/g. How much heat must be removed in changing 20g of this substance at 150 °C to -30 °C?
To solve this problem, we need to consider the different temperature ranges and phase changes of the substance. Let's break down the steps:
1. First, we need to calculate the amount of heat required to cool the substance from 150 °C to its freezing point of -10 °C as a liquid.
The heat formula is given by:
Q = m * c * ΔT
Where:
Q is the amount of heat transferred,
m is the mass of the substance,
c is the specific heat of the substance, and
ΔT is the change in temperature.
Using the specific heat of the liquid state (1.8 cal/g°C) and the mass (20g), the change in temperature is:
ΔT = -10°C - 150°C = -160°C
Now, we can calculate the heat required:
Q1 = 20g * 1.8 cal/g°C * -160°C
2. Next, we need to consider the phase change from a liquid at its freezing point (-10°C) to a solid at its freezing point (-10°C).
The heat required for the phase change from a liquid to a solid is called the heat of fusion, given as 100 cal/g. The mass remains the same (20g).
Q2 = 20g * 100 cal/g
3. After the substance has become a solid at -10°C, we need to calculate the amount of heat required to cool it further from -10°C to -30°C as a solid.
Using the specific heat of the solid state (0.8 cal/g°C) and the mass (20g), the change in temperature is:
ΔT = -30°C - (-10°C) = -30°C + 10°C = -20°C
Now, we can calculate the heat required:
Q3 = 20g * 0.8 cal/g°C * -20°C
4. Finally, we need to calculate the total amount of heat required by summing up the individual quantities Q1, Q2, and Q3:
Total amount of heat = Q1 + Q2 + Q3
Please note that we will consider the signs of these quantities (positive or negative) while adding them up.
Total amount of heat = Q1 + Q2 + Q3
= (20g * 1.8 cal/g°C * -160°C) + (20g * 100 cal/g) + (20g * 0.8 cal/g°C * -20°C)
To calculate the amount of heat that must be removed in changing the substance from 150 °C to -30 °C, we need to consider the different phase changes and temperature changes.
1. Phase Change from 150 °C to the boiling point at 140 °C (Gas to Liquid):
First, calculate the heat required to cool the substance from 150 °C to its boiling point at 140 °C using the specific heat as a gas:
Heat = mass x specific heat as a gas x temperature change
Heat = 20g x 0.3 cal/g.°C x (140 °C - 150 °C)
Heat = 20g x 0.3 cal/g.°C x (-10 °C)
Heat = -60 cal (Negative sign indicates heat removal)
2. Phase change from the boiling point at 140 °C to the freezing point at -10 °C (Liquid to Liquid):
Next, calculate the heat required to change the substance from a liquid at its boiling point to a liquid at its freezing point:
Heat = mass x heat of vaporization
Heat = 20g x 480 cal/g
Heat = 9600 cal (Positive sign indicates heat absorption)
3. Phase change from the freezing point at -10 °C to -30 °C (Solid to Solid):
Then, calculate the heat required to cool the substance from its freezing point to -30 °C using the specific heat as a solid:
Heat = mass x specific heat as a solid x temperature change
Heat = 20g x 0.8 cal/g.°C x (-30 °C - (-10 °C))
Heat = 20g x 0.8 cal/g.°C x (-20 °C)
Heat = -320 cal (Negative sign indicates heat removal)
Total heat removed = (-60 cal) + (9600 cal) + (-320 cal)
Total heat removed = 9220 cal