Change 80 degrees Celsius to Fahrenheit

Drew Williamson
2022-09-30
Answered

Change 80 degrees Celsius to Fahrenheit

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recreerefr

Answered 2022-10-01
Author has **7** answers

Temperature can be measured in various units.

Given temperature is 80 degree Celsius. In order to convert it into Fahrenheit, the following formula is used.

Celsius to Fahrenheit:

${(}^{\circ}C\times \frac{9}{5})+32{=}^{\circ}F\phantom{\rule{0ex}{0ex}}({80}^{\circ}C\times \frac{9}{5})+32={176}^{\circ}F$

The temperature in Fahrenheit for 80 degree Celsius is ${176}^{\circ}F$

Given temperature is 80 degree Celsius. In order to convert it into Fahrenheit, the following formula is used.

Celsius to Fahrenheit:

${(}^{\circ}C\times \frac{9}{5})+32{=}^{\circ}F\phantom{\rule{0ex}{0ex}}({80}^{\circ}C\times \frac{9}{5})+32={176}^{\circ}F$

The temperature in Fahrenheit for 80 degree Celsius is ${176}^{\circ}F$

asked 2022-04-26

Why the E of the time component 4-momentum is the total energy and not another?

The time component of the 4-momentum is $E/c$, and I saw that it is the "total energy" and from here we can derive the formula ${E}^{2}=(pc{)}^{2}+{m}^{2}{c}^{4}$

$E$ is the total energy? Can't it be some multiple of it or just some other energy?

The time component of the 4-momentum is $E/c$, and I saw that it is the "total energy" and from here we can derive the formula ${E}^{2}=(pc{)}^{2}+{m}^{2}{c}^{4}$

$E$ is the total energy? Can't it be some multiple of it or just some other energy?

asked 2022-10-15

A temperature of 20 C is equivalent to approximately

A. 6 F.

B. 32 F.

C. 68 F.

D. 136 F

A. 6 F.

B. 32 F.

C. 68 F.

D. 136 F

asked 2022-11-20

If 50 degrees Fahrenheit - what is the temperature in degrees Celsius and Kelvin (F = 1.8C +32)?

asked 2022-05-15

A Calorimetry Problem

I have a question in calorimetry from an old competitive exam. The question is:

The temperature of $100$ grams of water is to be raised from ${24}^{\circ}$C to ${90}^{\circ}$C by adding steam to it. Find the mass of steam required.

I tried attempting the question by assuming that the added steam would convert back to water and thus lose an amount of heat calculated by the latent heat of vaporization. Additionally, to cool from 100 degrees to 90 degrees, the 'watered steam' will contribute some additional heat. I equated this to the heat gained by the 100 grams of water to reach 90 degrees from 24 degrees. But I am not getting the right answer!

I would appreciate any help on this matter. Thank you :)

I have a question in calorimetry from an old competitive exam. The question is:

The temperature of $100$ grams of water is to be raised from ${24}^{\circ}$C to ${90}^{\circ}$C by adding steam to it. Find the mass of steam required.

I tried attempting the question by assuming that the added steam would convert back to water and thus lose an amount of heat calculated by the latent heat of vaporization. Additionally, to cool from 100 degrees to 90 degrees, the 'watered steam' will contribute some additional heat. I equated this to the heat gained by the 100 grams of water to reach 90 degrees from 24 degrees. But I am not getting the right answer!

I would appreciate any help on this matter. Thank you :)

asked 2022-05-09

Equilibrium temperature of closed system

Body X of temperature 0° C is brought into thermal contact with body Y of temperature 100° C. X has specific heat capacity higher than of Y. The masses of X and Y are equal.

By my reasoning, the final equilibrium temperature should lie between 0° C and 50° C. Is this correct?

Edit: 1) The bodies are in thermal contact only with one another; they are in a closed system.

2) My reasoning:

${Q}_{x}={m}_{x}{c}_{x}\mathrm{\Delta}{T}_{x}$

${Q}_{y}={m}_{y}{c}_{y}\mathrm{\Delta}{T}_{y}$

${Q}_{x}={Q}_{y}$, ${m}_{x}={m}_{y}$

${c}_{x}\mathrm{\Delta}{T}_{x}={c}_{y}\mathrm{\Delta}{T}_{y}$

If ${c}_{x}$ is higher than ${x}_{y}$, then $\mathrm{\Delta}{T}_{x}$ must be lower thab $\mathrm{\Delta}{T}_{y}$, so the equilibrium temperature must lie below 50° C.

Body X of temperature 0° C is brought into thermal contact with body Y of temperature 100° C. X has specific heat capacity higher than of Y. The masses of X and Y are equal.

By my reasoning, the final equilibrium temperature should lie between 0° C and 50° C. Is this correct?

Edit: 1) The bodies are in thermal contact only with one another; they are in a closed system.

2) My reasoning:

${Q}_{x}={m}_{x}{c}_{x}\mathrm{\Delta}{T}_{x}$

${Q}_{y}={m}_{y}{c}_{y}\mathrm{\Delta}{T}_{y}$

${Q}_{x}={Q}_{y}$, ${m}_{x}={m}_{y}$

${c}_{x}\mathrm{\Delta}{T}_{x}={c}_{y}\mathrm{\Delta}{T}_{y}$

If ${c}_{x}$ is higher than ${x}_{y}$, then $\mathrm{\Delta}{T}_{x}$ must be lower thab $\mathrm{\Delta}{T}_{y}$, so the equilibrium temperature must lie below 50° C.

asked 2022-05-14

Entropy change in a calorimetry problem

A standard textbook problem has us calculate the change in entropy in a system that undergoes some sort of heat exchange. For example, object $A$ has specific heat ${c}_{a}$ and initial temperature ${T}_{A}$ and object $B$ has specific heat ${c}_{b}$ with initial temperature ${T}_{B}$. They are they put in contact with each other until they reach thermal equilibrium, and our goal is to find the total entropy change of the system.

The standard solution is to use

$S=\int \frac{dQ}{T}$

where $dQ=mcdT$. But the above integral is only satisfied for reversible processes, whereas this heat exchange is clearly irreversible.

The usual workaround for this is to pick some reversible path and calculate the entropy change on our "fake" path, since entropy is a state variable. For example, in the free expansion of an ideal gas, we pick calculate the entropy change along an isotherm that carries us along the expansion to find the true change in entropy.

My question is - what exactly is the reversible path we are using when we use $dQ=mcdT$?

A standard textbook problem has us calculate the change in entropy in a system that undergoes some sort of heat exchange. For example, object $A$ has specific heat ${c}_{a}$ and initial temperature ${T}_{A}$ and object $B$ has specific heat ${c}_{b}$ with initial temperature ${T}_{B}$. They are they put in contact with each other until they reach thermal equilibrium, and our goal is to find the total entropy change of the system.

The standard solution is to use

$S=\int \frac{dQ}{T}$

where $dQ=mcdT$. But the above integral is only satisfied for reversible processes, whereas this heat exchange is clearly irreversible.

The usual workaround for this is to pick some reversible path and calculate the entropy change on our "fake" path, since entropy is a state variable. For example, in the free expansion of an ideal gas, we pick calculate the entropy change along an isotherm that carries us along the expansion to find the true change in entropy.

My question is - what exactly is the reversible path we are using when we use $dQ=mcdT$?

asked 2022-04-07

What calorimetry approach is this?

Question:

Find the final tempertature of the system when $5gm$ of steam at ${100}^{\circ}C$ is mized with $31.5gm$ ice at $-{20}^{\circ}C$

Constants:

Specific heat of ice and specific heat of steam is $0.5cal/gm{/}^{\circ}C$

Specific heat of water is $1cal/gm{/}^{\circ}C$

Latent heat of fusion is $80cal/gm$

Latent heat of vaporization is $540cal/gm$

My attempt:

I usually approach these problems by setting the final temperature as $T$. Then, I equate heat lost by steam to heat gained by ice. I have been taught this method and I am adept at it.

My problem: The solution given in my workbook is:

Heat released when $1g$ of steam at ${100}^{\circ}C$ is cooled to $-{20}^{\circ}C$ is $730cal$. Now, this heat is given to be given to $1+6.3g=7.3g$ of ice. This raises the temperature by ${10}^{\circ}C$

This is a new method for me, and I am unable to understand it. I wish to know:

How does this method work? In what calorimetry situations can this method be applied?

NOTE: The final answer is ${10}^{\circ}C$

Question:

Find the final tempertature of the system when $5gm$ of steam at ${100}^{\circ}C$ is mized with $31.5gm$ ice at $-{20}^{\circ}C$

Constants:

Specific heat of ice and specific heat of steam is $0.5cal/gm{/}^{\circ}C$

Specific heat of water is $1cal/gm{/}^{\circ}C$

Latent heat of fusion is $80cal/gm$

Latent heat of vaporization is $540cal/gm$

My attempt:

I usually approach these problems by setting the final temperature as $T$. Then, I equate heat lost by steam to heat gained by ice. I have been taught this method and I am adept at it.

My problem: The solution given in my workbook is:

Heat released when $1g$ of steam at ${100}^{\circ}C$ is cooled to $-{20}^{\circ}C$ is $730cal$. Now, this heat is given to be given to $1+6.3g=7.3g$ of ice. This raises the temperature by ${10}^{\circ}C$

This is a new method for me, and I am unable to understand it. I wish to know:

How does this method work? In what calorimetry situations can this method be applied?

NOTE: The final answer is ${10}^{\circ}C$