The figure below is a heat curve for an unknown outerspace substance discovered by the Mars rover. By understanding the heat curve, scientists can make inferences about the composition of the planet.
What is the temperature of fusion, Tfusion, and vaporization, Tvapor, for this unknown substance?
Y intercept is Temperature (C)
X intercept is Time
Start at (0,-60C)
then (20,80C) to (50,80C)
increase to (100,280C) to (140,280C)
increase to (160,360C)
For a sample of 67.4 g of this substance with a specific heat of 49.9 J/kg·C, how much heat is added to the substance if the temperature is to change from 180.0C to 250.0C?
Q= ? J
Q = (67.4 g)(49.9 J/kg·C)(250.0C - 180.0C)
Q = 5,845.2 J
To determine the temperature of fusion (Tfusion) and vaporization (Tvapor) for the unknown substance, we need to identify the points on the heat curve where changes in state occur.
From the given information, we can observe the following changes in temperature on the heat curve:
1. Start at (0, -60°C)
2. Increase to (20, 80°C) - This indicates a solid-to-liquid phase change (fusion).
3. Remain at (50, 80°C) - This indicates that the substance is in the liquid phase.
4. Increase to (100, 280°C) - This indicates a liquid-to-gas phase change (vaporization).
5. Remain at (140, 280°C) - This indicates that the substance is in the gaseous phase.
6. Increase to (160, 360°C) - This indicates another solid-to-liquid phase change.
7. The heat curve does not provide information about any further changes in state, so we cannot determine the temperature of vaporization (Tvapor).
Therefore, based on the given information, we can conclude that the temperature of fusion (Tfusion) is equal to 80°C, but we cannot determine the temperature of vaporization (Tvapor) from the provided data.
Moving on to the second part of the question, we need to calculate the amount of heat (Q) needed to change the temperature of a 67.4 g sample of the substance from 180.0°C to 250.0°C using the specific heat (c) provided.
The equation to calculate heat is:
Q = m * c * ΔT
Where:
Q = Heat energy (Joules)
m = Mass of the substance (67.4 g)
c = Specific heat capacity (49.9 J/kg·°C)
ΔT = Change in temperature (250.0°C - 180.0°C)
First, we need to convert the mass from grams to kilograms:
Mass (kg) = Mass (g) / 1000
Mass (kg) = 67.4 g / 1000
Mass (kg) = 0.0674 kg
Now we can calculate the heat:
Q = 0.0674 kg * 49.9 J/kg·°C * (250.0°C - 180.0°C)
Q = 0.0674 kg * 49.9 J/kg·°C * 70°C
Q ≈ 234.637 J (rounded to three decimal places)
Therefore, the amount of heat needed to change the temperature of the substance from 180.0°C to 250.0°C is approximately 234.637 J.
To find the temperature of fusion (Tfusion) and vaporization (Tvapor) for the unknown substance, we need to look for any plateaus or flat regions in the heat curve.
From the given heat curve, we can observe two such regions:
1. The first flat region occurs between 0°C and 80°C. This represents the temperature of fusion (Tfusion) where the substance is changing phase from solid to liquid.
2. The second flat region occurs between 100°C and 280°C. This represents the temperature of vaporization (Tvapor) where the substance is changing phase from liquid to gas.
Therefore, the temperature of fusion (Tfusion) is 80°C and the temperature of vaporization (Tvapor) is 280°C.
Now, to calculate the amount of heat added to the substance when the temperature changes from 180.0°C to 250.0°C, we can use the formula:
Q = mcΔT
Where:
Q is the heat energy added (in Joules)
m is the mass of the substance (in kg)
c is the specific heat capacity of the substance (in J/kg·°C)
ΔT is the change in temperature (in °C)
Given:
Mass of the substance (m) = 67.4 g = 0.0674 kg
Specific heat capacity (c) = 49.9 J/kg·°C
Change in temperature (ΔT) = 250.0°C - 180.0°C = 70.0°C
Plugging these values into the formula, we get:
Q = (0.0674 kg)(49.9 J/kg·°C)(70.0°C)
Q = 235.0686 J
Therefore, the amount of heat added to the substance is approximately 235.07 J.