Answer to Question #95517 in Molecular Physics | Thermodynamics for fatmat

Question

a faulty mercury thermometer reads 10 degrees celcius when dipped into melting ice and 90 degrees celcius when on steam at normal atmospheric pressure. what would this thermometer read when dipped into a liquid at 20 degrees celcius

Answer 1

Solution:

Let's use the formula for linear thermal expansion:

$\Delta{}l=l_0\alpha\Delta{}t$

Let's write this formula for l₀=10, l₁=90, t₀=0^oC, t₁=100^oC:

$l_1-l_0=l_0\alpha(t_1-t_0)$

We can find linear thermal coefficient of mercury:

$\alpha=\frac{l_1-l_0}{l_0(t_1-t_0)}=\frac{90-10}{10\cdot{}(100-0)}=0.08$ K^-1.

Now let's write the formula for l₀=10, t₀=0^oC, t₂=20^oC:

$l_2-l_0=l_0\alpha(t_2-t_0)$

$l_2=l_0+l_0\alpha(t_2-t_0)=$

$=10+10\cdot{}0.08\cdot{}(20-0)=26$

Answer:

Thermometer will read 26^o C.