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(a) The temperature on a warm summer day is \(87^{\circ} \mathrm{F}\) . What is the temperature in \(^{\circ} \mathrm{C} ?\) (b) Many scientific data are reported at \(25^{\circ} \mathrm{C}\) . What is this temperature in kelvins and in degrees Fahrenheit? (c) Suppose that a recipe calls for an oven temperature of \(400^{\circ} \mathrm{F}\) . Convert this temperature to degrees Celsius and to kelvins. (d) Liquid nitrogen boils at 77 \(\mathrm{K}\) . Convert this temperature to degrees Fahrenheit and to degrees Celsius.

Short Answer

Expert verified
(a) 87°F is approximately \(30.56^{\circ} \mathrm{C}\). (b) 25°C is equivalent to 298.15K and \(77^{\circ} \mathrm{F}\). (c) 400°F is approximately \(204.44^{\circ} \mathrm{C}\) and 477.59K. (d) 77K is equivalent to \(-196.15^{\circ} \mathrm{C}\) and approximately \(-320.87^{\circ} \mathrm{F}\).

Step by step solution

01

(a) Convert 87°F to Celsius

First, apply the Fahrenheit to Celsius formula: \(C = (F - 32) \times \frac{5}{9}\) Substitute F with 87: \(C = (87 - 32) \times \frac{5}{9}\) Now, calculate the Celsius temperature: \(C = 55 \times \frac{5}{9} \approx 30.56^{\circ}\)
02

(b) Convert 25°C to Kelvin and Fahrenheit

Converting Celsius to Kelvin: \(K = C + 273.15\) Substitute C with 25: \(K = 25 + 273.15 = 298.15K\) Converting Celsius to Fahrenheit: \(F = (C \times \frac{9}{5}) + 32\) Substitute C with 25: \(F = (25 \times \frac{9}{5}) + 32 \approx 77^{\circ}\)
03

(c) Convert 400°F to Celsius and Kelvin

Converting Fahrenheit to Celsius: \(C = (F - 32) \times \frac{5}{9}\) Substitute F with 400: \(C = (400 - 32) \times \frac{5}{9} \approx 204.44^{\circ}\) Now convert Celsius to Kelvin: \(K = C + 273.15\) Substitute C with 204.44: \(K = 204.44 + 273.15 = 477.59K\)
04

(d) Convert 77K to Fahrenheit and Celsius

Converting Kelvin to Celsius: \(C = K - 273.15\) Substitute K with 77: \(C = 77 - 273.15 = -196.15^{\circ}\) Now convert Celsius to Fahrenheit: \(F = (C \times \frac{9}{5}) + 32\) Substitute C with -196.15: \(F = (-196.15 \times \frac{9}{5}) + 32 \approx -320.87^{\circ}\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Celsius to Fahrenheit
Temperature conversion from Celsius to Fahrenheit is a simple yet important concept. When converting from Celsius to Fahrenheit, you multiply the Celsius temperature by \(\frac{9}{5}\), then add 32 to the result. This formula helps determine how warm or cold a temperature in Celsius would feel in Fahrenheit.

Here is the formula in a straightforward way:
  • Start with your Celsius temperature.
  • Multiply by 9, then divide by 5.
  • Add 32. This will give you the temperature in Fahrenheit.
For example, converting \(25^{\circ} \mathrm{C}\) to Fahrenheit would look like this:- Multiply 25 by \(9/5\), which equals 45.- Add 32, resulting in \(77^{\circ} \mathrm{F}\).

This conversion is useful for everything from understanding the weather in different countries to cooking recipes that use Fahrenheit.
Fahrenheit to Celsius
The process of converting Fahrenheit to Celsius is the reverse of converting Celsius to Fahrenheit. You start by subtracting 32 from the Fahrenheit temperature. This is because the two scales have different starting points for freezing. Next, you multiply the result by \(\frac{5}{9}\) to adjust for the scale difference.

The formula for converting Fahrenheit to Celsius is:
  • Subtract 32 from the Fahrenheit temperature.
  • Multiply the result by \(\frac{5}{9}\).
Consider converting \(87^{\circ} \mathrm{F}\) to Celsius:- Subtract 32 from 87, which gives 55.- Multiply 55 by \(\frac{5}{9}\) to get approximately \(30.56^{\circ} \mathrm{C}\).

This conversion is especially useful in scientific contexts where Celsius is more commonly used.
Celsius to Kelvin
Converting Celsius to Kelvin is quite intuitive since both scales increment linearly, meaning each degree change in Celsius is equal to a degree change in Kelvin. The only adjustment needed is accounted by the offset factor between the two, which is 273.15. The Kelvin scale starts at absolute zero, which is the lowest possible temperature.

Follow these steps for the conversion:
  • Add 273.15 to the Celsius temperature.
For instance, converting \(25^{\circ} \mathrm{C}\) to Kelvin involves simply adding 273.15, arriving at \(298.15 \mathrm{K}\).

This method is particularly crucial in scientific experiments where temperatures in Kelvin are often required for calculations involving gas laws and thermodynamics.
Kelvin to Fahrenheit
To convert Kelvin to Fahrenheit, two steps are needed since there isn't a direct conversion formula. First, convert Kelvin to Celsius by subtracting 273.15. Then, use the Celsius to Fahrenheit formula to complete the conversion.

Here's the two-step process:
  • Convert Kelvin to Celsius: subtract 273.15 from the Kelvin temperature.
  • Convert Celsius to Fahrenheit: multiply the Celsius temperature by \(\frac{9}{5}\) and add 32.
For example, to convert \(77 \mathrm{K}\) to Fahrenheit:- First, convert to Celsius: \(77 - 273.15 = -196.15^{\circ} \mathrm{C}\).- Then convert to Fahrenheit: \(-196.15 \times \frac{9}{5} + 32 \approx -320.87^{\circ} \mathrm{F}\).

This conversion is helpful in fields like cryogenics where Kelvin is commonly used, and understanding temperatures in Fahrenheit can provide more context.

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Most popular questions from this chapter

(a) After the label fell off a bottle containing a clear liquid believed to be benzene, a chemist measured the density of the liquid to verify its identity. A \(25.0-\) mL portion of the liquid had a mass of 21.95 \(\mathrm{g} .\) A chemistry handbook lists the density of benzene at \(15^{\circ} \mathrm{C}\) as 0.8787 \(\mathrm{g} / \mathrm{mL} .\) Is the calculated density of benzene at \(15^{\circ} \mathrm{C}\) as 0.8787 \(\mathrm{g} / \mathrm{mL} .\) Is the calculated density in agreement with the tabulated value? (b) An experiment requires 15.0 \(\mathrm{g}\) of cyclohexane, whose density at \(25^{\circ} \mathrm{C}\) is 0.7781 \(\mathrm{g} / \mathrm{mL}\) . What volume of cyclohexane should be used? (c) A spherical ball of lead has a diameter of 5.0 \(\mathrm{cm} .\) What is the mass of the sphere if lead has a density of 11.34 \(\mathrm{g} / \mathrm{cm}^{3} ?\) (The volume of a sphere is \((4 / 3) \pi r^{3},\) where \(r\) is the radius.)

Gold is alloyed (mixed) with other metals to increase its hardness in making jewelry. (a) Consider a piece of gold jewelry that weighs 9.85 \(\mathrm{g}\) gond has a volume of 0.675 \(\mathrm{cm}^{3} .\) The jewelry contains only gold and silver, which have densities of 19.3 and 10.5 \(\mathrm{g} / \mathrm{cm}^{3}\) , respectively. If the total volume of the jewelry is the sum of the volumes of the gold and silver that it contains, calculate the percentage of gold (by mass) in the jewelry. (b) The relative amount of gold in an alloy is commonly expressed in units of carats. Pure gold is 24 carat, and the percentage of gold in an alloy is given as a percentage of this value. For example, an alloy that is 50\(\%\) gold is 12 carat. State the purity of the gold jewelry in carats.

For each of the following processes, does the potential energy of the object(s) increase or decrease? (a) The distance between two oppositely charged particles is increased. (b) Water is pumped from ground level to the reservoir of a water tower 30 \(\mathrm{m}\) above the ground. (c) The bond in a chlorine molecule, \(\mathrm{Cl}_{2},\) is broken to form two chlorine atoms.

Using your knowledge of metric units, English units, and the information on the back inside cover, write down the con- version factors needed to convert (a) mm to nm, (b) mg to kg, (c) km to ft, (d) in. \(^{3}\) to \(\mathrm{cm}^{3} .\)

Gold can be hammered into extremely thin sheets called gold leaf. An architect wants to cover a 100 \(\mathrm{ft} \times 82\) ft ceiling with gold leaf that is five-millionths of an inch thick. The density of gold is \(19.32 \mathrm{g} / \mathrm{cm}^{3},\) and gold costs \(\$ 1654\) per troy ounce \((1\) troy ounce \(=31.1034768 \mathrm{g}) .\) How much will it cost the architect to buy the necessary gold?

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