Question

Avalanche forecasters measure the temperature gradient , which is the rate at which the temperature in a snowpack T changes with respect to its depth h. A large temperature gradient may lead to a weak layer in the snowpack. When these weak layers collapse, avalanches occur. Avalanche forecasters use the rule of thumb that if exceeds 10°C/m anywhere in the snowpack, conditions are favorable for weak-layer formation, and the risk of avalanche increases. Assume the temperature function is continuous and differentiable. Complete parts (a) through (d) below. a. An avalanche forecaster digs a snow pit and takes two temperature measurements. At the surface (h=0), the temperature is -11°c. At a depth of 1.8 m, the temperature is -5°C. Using the Mean Value The average temperature gradient from h=0 to h = 1.8 is lºc/m. (Simplify your answer. Round to the nearest tenth as needed.) Is the formation of a weak layer likely? Select the correct answer below. O Ο Yes Νο b. One mile away, a skier finds that the temperature at a depth of 1.1 m is -1°c, and at the surface it is - 14°C. What can be concluded about the temperature gradient? The average temperature gradient from h = 0 to h= 1.1 is lºc/m. (Simplify your answer. Round to the nearest tenth as needed.)

Answer 1

Answers:-

(a) 3.3℃/ meter , no weak layer formed

(b)11.8℃/meter, yes,weak layer is formed

(c)correct option-D

(d)correct option-C

Solution-

Temperature gradient is defined as the rate of change of temperature(T) in the snowpack with respect to depth h.So, average temperature gradient is

$\frac{dT}{dh}=\frac{T'-T_o}{h}$ ............1

Here,

To is the temperature at surface

T' is the temperature at depth h meters

h is the depth in meters.

Now,

(a) According to question,

Temperature at surface(h=0) is To= -11℃

Temperature at the depth h is T' = -5℃

Depth is h=1.8 m

Now on putting these values in equation 1, we get, the average temperature gradient as

$\frac{dT}{dh}=\frac{-5-(-11)}{1.8}$

$\frac{dT}{dh}=\frac{-5+11}{1.8}$

$\frac{dT}{dh}=\frac{6}{1.8}$

$\frac{dT}{dh}=3.3^{\circ}C/m$

Since, average temperature gradient is less than 10℃/m .So, no weak layer will formed.

(b) According to question,

One miles away

Temperature at surface(h=0) is To= -14℃

Temperature at the depth h is T' = -1℃

Depth is h=1.1 m

Now on putting these values in equation 1, we get, the average temperature gradient as

$\frac{dT}{dh}=\frac{-1-(-14)}{1.1}$

$\frac{dT}{dh}=\frac{-1+14}{1.1}$

$\frac{dT}{dh}=\frac{13}{1.1}$

$\frac{dT}{dh}=11.8^{\circ}C/m$

Since, average temperature gradient is more than 10℃/m .So, here weak layer will be formed.

(c)

We can observe from part (b),weak layer is more likely to form at places where snowpack gets thinner, the temperature gradient is larger because temperature gradient is in inverse relation with the depth.

Since

$\frac{dT}{dh}=\frac{T'-T_o}{h}$

So, as h ↓(decreses) then dT/dh ↑(increases)

Hence, the correct option is (D).

(d) Isothermal means temperature is constant.

So, if T'=To , than temperature gradient is

$\frac{dT}{dh}=\frac{0-0}{h}=0$

It is smaller than 10℃/m .

Hence, the correct option is (C).No,weak layer will form in isothermal snow.If all the layers are at same temperature, than temperature difference is 0.So, temperature gradient is 0.