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mojhsa [17]
3 years ago
11

If the temperature changes by 100 K, by how much does it change in degrees Celsius?

Chemistry
1 answer:
Bumek [7]3 years ago
4 0

Taking into account the degrees Celsius and Kelvin, if the temperature changes by 100 K, by 100 degrees Celsius it change in degrees Celsius.

A degree Centigrade is a unit of temperature (the physical quantity that reveals the level of heat) whose symbol is °C.

On the other hand, the degree Kelvin, whose symbol is K, is the unit of temperature on the scale created by William Thomson Kelvin, in 1848, based on the degree Celsius, he established his zero point (0) at zero absolute (-273.15 ° C). The figure 0 K is equivalent to the theoretical point where gases have no volume. Absolute zero or 0 K is the point where molecules theoretically stop moving. It is the "perfect" cold.

Considering that zero on the Celsius or degrees centigrade (0 ° C) scale is defined as the equivalent of 273.15 K, the conversion from degrees ° C to K is carried out using the following expression:

T(K)= T(°C) + 273.15 K

In this case, if the temperature changes by 100 K, as it is a change in temperature, the numerical value in K units is the same in ° C units.

Then, the correct answer is option B. if the temperature changes by 100 K, by 100 degrees Celsius it change in degrees Celsius.

Learn more about degrees Celsius and Kelvin:

  • brainly.com/question/2800860?referrer=searchResults
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Julli [10]

Well, first we must remember that

pK_{a}+pK_{b}=14

This is because

K_{a}*K_{b}=10^{-14}

-log(K_{a}*K_{b})=-log(10^{-14})\\-logK_{a}+-logK_{b}=-log(10^{-14})\\pK_{a}+pK_{b}=14

So then

pK_{b}=14-pK_{a}=14-3.2=1.8

7 0
4 years ago
Read 2 more answers
An iron block of mass 18 kg is heated from 285 K to 318 K. If 267.3 kJ is required, what is the specific heat of iron?
creativ13 [48]

Answer:

D. 450 J/kgK

Explanation:

Using the formula as follows:

Q = m × c × ∆T

Where:

Q = amount of heat absorbed/released (J)

m = mass of substance (g)

c = specific heat capacity

∆T = change in temperature (°C)

According to the information provided in this question:

Q = 267.3 kJ = 267300J

m = 18kg

∆T = 318K - 285K = 33K

c = ?

Q = m × c × ∆T

c = Q ÷ m∆T

c = 267300 ÷ 18 × 33

c = 267300 ÷ 594

c = 450 J/kgK

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chubhunter [2.5K]

Answer:

c. Li_2CO_3

Explanation:

Hello,

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Answer:

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3 years ago
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