Typical temperatures in the interior of the Earth and Sun are about 4000°C and 15 x 10 6 °C, respectively. What percent error is made in each case if a person forgets to change °C to K?

Welcome back everyone in this problem in the course of Jupiter and a white dwarf star, the typical temperatures are about three multiplied by 10 to the fourth degree Celsius and five multiplied by 10 to the sixth degree Celsius respectively. If someone forgets to change degrees Celsius to Kelvin, what percent error is made in each case? A says that the percentage error for Jupiter is 0.1%. While for the white dwarf, it's 0.01%. B says they are 0.9% and 0.005% respectively. C says they are one and 0.1% respectively. And D says they are 10 and 1% respectively. Now, if we're gonna figure out a percentage error from forgetting to change from degrees Celsius to Kelvin, it would be good to have our temperatures in Kelvin for comparison. OK. Now recall that to convert a temperature from degrees Celsius to Kelvin. OK. It basically means that we would add 273.15 to our temperature in degrees Celsius. OK. So applying that idea then that means for Jupiter. Yeah, for Jupiter with a temperature of three multiplied by 10 to the fourth degree Celsius. Its temperature in Kelvin is gonna be equal to that volume, OK? In degrees Celsius plus 273.15 which is gonna be equal to 30,273 0.15. Kelvin. Likewise for the white dwarf star, OK, then its temperature in Kelvin is going to be equal to its temperature in Celsius. Five multiplied by 10 to the sixth degrees Celsius plus 273.15 which is going to be equal to 5,000,273 0.15. Kelvin. Ok. No, if that's the case now that we know their temperatures in Kelvin, we can test out our percentage error. Now how are we going to do that? Well, the percentage error is going to be equal to and let me write that here that percentage error is going to be equal to the temperature change in this case or 273.15 divided by our temperature in Kelvin multiplied by 100. When we multiply it by 100 it turns it into a percentage. So when we're talking about our percentage error for Jupiter, then, OK, then Jupiter's percentage error is going to be equal to 273.15 K. Kelvin divided by its temperature in Kelvin, 30,273 0.15. Kevin multiplied by 100 which is equal to 0.90228%. Or if we run that off, it's approximately equal to 0.9%. Ok. Likewise, for our white dwarfs error, then it's going to be equal to 273.15. Kelvin divided by 5 million, 273.15. Kelvin, which already we can tell the percentage error is going to be much smaller and we're multiplying it by 100. And that gives us 0.00546% or zero, approximately 0.005%. Therefore, the percentage error for Jupiter is 0.9%. And for our White Dwarf star, 0.005% which when we go back to our answer choices would be BB is the correct answer. Thanks for watching everyone. I hope this video helped.

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Problem 33b

Textbook Question

Typical temperatures in the interior of the Earth and Sun are about 4000°C and 15 x 10⁶ °C, respectively. What percent error is made in each case if a person forgets to change °C to K?

Verified step by step guidance

Convert the given temperatures from Celsius to Kelvin using the formula:

T

_K = T_°C + 273.15. For the Earth, the temperature in Kelvin is

T

_K,Earth = 4000 + 273.15. For the Sun, the temperature in Kelvin is

T

_K,Sun = 15 × 10⁶ + 273.15.

Understand the concept of percent error. Percent error is calculated using the formula:

\text{Percent Error} = \left| \frac{\text{True Value} - \text{Approximate Value}}{\text{True Value}} \right| × 100

. Here, the true value is the temperature in Kelvin, and the approximate value is the temperature in Celsius (since the person forgot to convert).

For the Earth, calculate the percent error. Substitute the true value as

T

_K,Earth and the approximate value as

T

_°C,Earth = 4000 into the percent error formula.

For the Sun, calculate the percent error. Substitute the true value as

T

_K,Sun and the approximate value as

T

_°C,Sun = 15 × 10⁶ into the percent error formula.

Compare the percent errors for the Earth and the Sun. Note that the percent error will be much smaller for the Sun because the added 273.15 K is negligible compared to the large temperature value, whereas for the Earth, the added 273.15 K is more significant relative to the temperature.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Temperature Scales

Temperature can be measured in different scales, primarily Celsius (°C) and Kelvin (K). The Kelvin scale is an absolute temperature scale where 0 K represents absolute zero, the point at which molecular motion ceases. To convert Celsius to Kelvin, one must add 273.15 to the Celsius temperature. Understanding this conversion is crucial for accurately calculating temperature-related physical properties.

Percent Error Calculation

Percent error is a way to express the accuracy of a measurement compared to the true value. It is calculated using the formula: Percent Error = |(Experimental Value - True Value) / True Value| × 100%. This concept is essential for evaluating how significant the error is when using Celsius instead of Kelvin, particularly in scientific contexts where precision is critical.

Thermal Energy and Its Implications

Thermal energy is the energy that comes from the temperature of matter, which is directly related to the motion of particles. In the context of the Earth and Sun, understanding thermal energy is important because it influences geological and astrophysical processes. The difference in temperature scales can lead to significant implications in calculations involving energy transfer, radiation, and other thermal dynamics.

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Typical temperatures in the interior of the Earth and Sun are about 4000°C and 15 x 106 °C, respectively. What percent error is made in each case if a person forgets to change °C to K?

Key Concepts

Temperature Scales

Percent Error Calculation

Thermal Energy and Its Implications

Watch next

Typical temperatures in the interior of the Earth and Sun are about 4000°C and 15 x 10⁶ °C, respectively. What percent error is made in each case if a person forgets to change °C to K?