Physics
1st Edition
ISBN: 9780133256925
Table of contents
Textbook solutions
All Solutions
Page 349: Practice Problems
Exercise 7
Solution 1
Solution 2
Step 1
1 of 2
(a) $6000 mathrm{K}=(6000-273)text{textdegree}mathrm{C}=5727text{textdegree}mathrm{C}$
Step 2
2 of 2
(b)
$$
T_F=frac{9}{5}T_C+32=frac{9}{5}times 5727+32=10340.6text{textdegree}mathrm{F}
$$
Step 1
1 of 3
a) To convert the temperature from Kelvin to Celsius, we use the following formula:
$$
T_C = T – 273.15
$$
Where $T = 6000text{ K}$,
$$
begin{align*}
T_C &= 6000 – 273.15\
\
T_C &= boxed{5726.85text{ }^circ text{C}}\
end{align*}
$$
Step 2
2 of 3
b) To convert the temperature from Celsius to Fahrenheit, we use the following formula:
$$
begin{align*}
T_F &= dfrac{9}{5}left( {T_C} right) + 32\
\
T_F &= dfrac{9}{5}left( {5726.85} right) + 32\
\
T_F &= 1.8left( {5726.85} right) + 32\
\
T_F &= 10308.33 + 32\
\
T_F &= boxed{10340.33text{ }^circ text{F}}\
end{align*}
$$
Result
3 of 3
$T_C = 5726.85text{ }^circ text{C}$
$$
T_F = 10340.33text{ }^circ text{F}
$$
Exercise 8
Step 1
1 of 2
We know that the boiling point of water is $100text{textdegree} C$
In Kelvin we have:
$$
begin{equation*}
T = T_c + 273.15 = 100 + 273.15 = 373.15 K
end{equation*}
$$
Result
2 of 2
$$
T = 373.15 K
$$
T = 373.15 K
$$
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