Physics, 05.08.2019 16:10 ayoismeisalex

16 grams of ice at –32°c is to be changed to steam at 182°c. the entire process requires cal. round your answer to the nearest whole number. the specific heat of both ice and steam is 0.5 cal/g°c. the specific heat of water is 1.00 cal/gk. the heat of fusion is 80 cal/g and the heat of vaporization is 540 cal/g.

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Answer from: darkmayem

12432 cal.

Explanation:

The process to change ice at -32 ºC to steam at 182 ºC can be divided into 5 steps:

1. Heat the ice to 0 ºC, which is the fusion temperature.

2. Melt the ice (obtaining liquid water), which is a process at constant pressure and temperature, so the liquid obtained is also at 0ºC.

3. Heat the liquid water from 0 ºC to 100 ºC, which is the vaporization normal temperature of the water.

4. Vaporization of all the water; this is also a process that occurs at constant pressure and temperature, so the produced steam will be at 100ºC.

5. Heat the steam from 100 ºC to 182 ºC.

Each process has a required energy, and the sum of the energy required for each and all of the steps is the total amount of energy required for the whole process:

E_T=E_1+E_2+E_3+E_4+E_5

E_1 is a heating process for the ice, so we know that the energy required is proportional to the temperature difference through the specific heat:

$E_1=m*Cp_{sol}*(T_2-T_1)\\E_1=16g*0.5\frac{cal}{gC}*(0-(-32))=256cal$

E_2 is a phase change process, so we do not use the specific heat (sensible heat), but the fusion heat (latent heat), so:

$E_{2}=m*dh_{f}={16g*80\frac{cal}{g}}=1280cal$

Analogously,

$E_3=m*Cp_{liq}*(T_3-T_2)=16g*1.00\frac{cal}{gC}*(100-0)C = 1600 cal$

$E_{4}=m*{dh_{vap}}\\\\E_4=16g*540\frac{cal}{g} =8640cal$

$E_{5}=m*Cp_{vap}*(T_{5}-T_{4})\\E_{5}={16g*0.5\frac{cal}{gK}*(182-100)K}=656cal$

Finally, the total energy required is:

$E_T=256cal+1280cal+1600cal+8640cal+656cal\\E_T=12432cal$

Answer from: Quest

colorectal cancer may also develop from areas of abnormal cells in the lining of the colon or rectum. this area of abnormal cells is called dysplasia and is more commonly seen in people with certain inflammatory diseases of the bowel such as crohn's disease or ulcerative colitis.

Answer from: Quest

we have been given a quadratic function $f(x)=(x-5)^{2} +2$ and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is $[2,\infty)$ .

if we restrict the domain of this function to either $(-\infty,5]$ or $[5,\infty)$ , it will become one to one function.

let us know find its inverse.

$y=(x-5)^{2}+2$

upon interchanging x and y, we get:

$x=(y-5)^{2}+2$

let us now solve this function for y.

$(y-5)^{2}=x-2\\ y-5=\pm \sqrt{x-2}\\ y=5\pm \sqrt{x-2}\\$

hence, the inverse function would be $f^{-1}(x)=5+\sqrt{x-2}$ if we restrict the domain of original function to $[5,\infty)$ and the inverse function would be $f^{-1}(x)=5-\sqrt{x-2}$ if we restrict the domain to $(-\infty,5]$ .

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16 grams of ice at –32°c is to be changed to steam at 182°c. the entire process requires cal. round...

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