Problem 1
a) The ambient temperature is usually take as 68ºF. How much is this on the celsius scale?
b)The temperature of a filament in a focus is 1900ºC, how much is this in ºF scale?
For this we use the following formula:
\(T\) (ºC)\(=\frac{5}{9}\)\(\left[T\left(F\right)-32\right]\)
\(T\) (ºF)\(=\frac{9}{5}\)\(\left[T\left(C\right)+32\right]\)
And we replace it
\(T\) (ºC)\(=\frac{5}{9}\)\(\left[68-32\right]=20\)
Then we take out in ºF
\(T\) (ºF)\(=\frac{9}{5}\)\(\left[1900\ C+32\right]=3478\)º
Problem 2
A thermometer in an alcohol is the alcohol column having a length of 11.82cm at ºC and a length of 21.85cm at 100ºC
a) What is the temperature if the column has a lngth of 14.60cm?
b)What is the temperature if the column has a length of 14.60cm?
For that we use the formula:
\(y=mx+b\)
How are you multiplying
\(m=\frac{y2-y1}{x2-x1}=\frac{100}{10.03}=9.97\)
We fit the values of the following form
\(0=\left(9.97\right)\left(11.82\right)+b\)
\(b=-\left(9.97\right)\left(11.82\right)=-117.84\)
Which is the value of
\(y=9.97x-117.84\)
\(=18.70\)
\(y=\left(9.97\right)\left(18.7\right)-117.84\ =68.59\)ºC
\(x=14.6\)
\(y=\left(9.97\right)\left(14.60\right)-117.84\ =27.71\)ºC
Problem 3
An average active person consumes about 2500 Kcalories a day.
a)How much is this in J.
b)In how many Kwats per hour
c)If CFE charges 2 sets per Kwats per hour, how much would the energy cost x/d if you buy it from CFE and you could feed this amount of money per day.
The formula is
w=mgh
a) \(\left(2500\right)\left(4.186x10^3J\right)\)
b) \(\frac{1.0465x10^6}{3.6x10^6}=2.9kwh\)
c)\(\frac{10\ c}{1\ kwh}=\ 29\ x\ 24hors\)
\(=6.96\ USD\)
Problem 4
How much J and Kcalories are generated when applying the brakes to stop a 1200kg car coming at 95km/h?
The formula we will use here will be:
\(Δk=kF-kI\)
We substitute the formula
\(=-\frac{1}{2}\left(1200kg\right)\left(95\ km\ por\ h\right)^2\)
\(=0-\left(600kg\right)\left(26.4\ \frac{m}{s}\right)^2\)
\(=-418.176J\)
\(=-418.176\ \frac{k}{J}\left(\frac{1\ Kcal}{4.186kJ}\right)\)\(=-99.9Kcal\)
Problem 5
The cooling system of an automobile has 18 liters of water that heat so much heat observed if its temperature is 15 to 95ºC?
The formula for this solution is:
\(Q=mcΔT\)
\(m=18kg\)
\(c=4186k\)ºc
\(ΔT=95\)º\(-15\)º=80º
This is equal to
\(\left(18\right)\left(4186\right)=\left(80\ grdsC\right)=6027x10^3\)
\(=6.0x10^6J\)