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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

$$
T = 373.15 K
$$

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Page 349: Practice Problems

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