On a day when the temperature reaches 53 degree F, what is the temperature in degrees Celsius and in kelvins? Conceptualize: In the United States, a temperature of 53 degree F is well understood. In many other parts of the world, however, this temperature might be meaningless because people are familiar with the Celsius temperature scale. Categorize: This example is a simple substitution problem. Solve: T_F = 9/5 T_C + 32 degree F for the Celsius temperature and substitute numerical values: T_C = 5/9 (T_F – 32) = 5/9 (53 – 32) = 11.66 degree C Use: T_C = T – 273.15 to find the Kelvin temperature: T = T_C + 273.15 = 284.82 K A convenient set of weather-related temperature equivalents to keep in mind is that 0 degree C is (literally) freezing at 32 degree F, 10 degree C is cool at 50 degree F, 20 degree C is room temperature, 30 degree C is warm at 86 degree F, and 40 degree C is a hot day at 104 degree F. A substance has a melting point of 1,243 degree C and a boiling point of 2,640 degree C. (a) Express these temperatures in kelvins. melting point = K boiling point = K (b) Express these temperatures in degrees Fahrenheit. melting point = degree F boiling point = degree F
The Correct Answer and Explanation is:
Let’s solve this step-by-step with detailed explanations.
Given:
- Temperature in Fahrenheit: TF=53∘FT_F = 53^\circ F
- Melting point: 1243∘C1243^\circ C
- Boiling point: 2640∘C2640^\circ C
Part 1: Convert 53°F to Celsius and Kelvin
The formula to convert Fahrenheit (TFT_F) to Celsius (TCT_C) is: TC=59(TF−32)T_C = \frac{5}{9} (T_F – 32)
Substituting TF=53∘FT_F = 53^\circ F: TC=59(53−32)=59×21=11.66∘CT_C = \frac{5}{9} (53 – 32) = \frac{5}{9} \times 21 = 11.66^\circ C
Now, to convert Celsius to Kelvin (TT), use: T=TC+273.15T = T_C + 273.15 T=11.66+273.15=284.81 KT = 11.66 + 273.15 = 284.81 \text{ K}
Part 2: Convert melting and boiling points from Celsius to Kelvin
Kelvin temperature is given by: K=∘C+273.15K = ^\circ C + 273.15
- Melting point:
1243+273.15=1516.15 K1243 + 273.15 = 1516.15 \text{ K}
- Boiling point:
2640+273.15=2913.15 K2640 + 273.15 = 2913.15 \text{ K}
Part 3: Convert melting and boiling points from Celsius to Fahrenheit
The formula to convert Celsius to Fahrenheit is: TF=95TC+32T_F = \frac{9}{5} T_C + 32
- Melting point:
TF=95×1243+32=2237.4+32=2269.4∘FT_F = \frac{9}{5} \times 1243 + 32 = 2237.4 + 32 = 2269.4^\circ F
- Boiling point:
TF=95×2640+32=4752+32=4784∘FT_F = \frac{9}{5} \times 2640 + 32 = 4752 + 32 = 4784^\circ F
Final Answers:
- 53°F = 11.66°C = 284.81 K
- Melting point = 1516.15 K = 2269.4°F
- Boiling point = 2913.15 K = 4784°F
Explanation (300+ words):
Temperature scales are used worldwide to measure thermal energy, but the familiarity and use of these scales vary. In the United States, the Fahrenheit scale is common, whereas most other countries use Celsius or Kelvin. Understanding how to convert between these scales is important in science and everyday life.
The Fahrenheit to Celsius conversion is a linear transformation based on the difference in the zero points and unit sizes between the two scales. The Celsius scale sets the freezing point of water at 0°C and boiling at 100°C, while Fahrenheit sets freezing at 32°F and boiling at 212°F. The formula TC=59(TF−32)T_C = \frac{5}{9}(T_F – 32) adjusts for these differences. For example, 53°F, while commonly understood in the U.S., corresponds to approximately 11.66°C in the Celsius scale, which is considered a cool temperature.
The Kelvin scale, on the other hand, is an absolute temperature scale used primarily in scientific contexts. It begins at absolute zero, the theoretical lowest possible temperature where molecular motion stops, defined as 0 K. The Kelvin scale increments are equal to Celsius degrees, so conversion is straightforward by adding 273.15. Thus, 11.66°C equals 284.81 K.
For substances with very high melting and boiling points, converting Celsius to Kelvin involves the same addition of 273.15. For example, a substance melting at 1243°C corresponds to 1516.15 K. The boiling point at 2640°C equals 2913.15 K. When converting these values to Fahrenheit, the formula TF=95TC+32T_F = \frac{9}{5} T_C + 32 expands the scale and shifts the zero point, resulting in very high Fahrenheit values (2269.4°F and 4784°F).
In summary, understanding these conversions is essential for cross-cultural communication of weather conditions, scientific data interpretation, and technical applications. The use of a common baseline like Kelvin aids scientific precision, while Celsius and Fahrenheit serve practical everyday needs in different parts of the world.