To solve this problem it is necessary to apply the concepts related to the Stefan-Boltzman law that is responsible for calculating radioactive energy.

Mathematically this expression can be given as

$P = \sigma Ae\Delta T^4$

Where

A = Surface area of the Object

$\sigma =$ Stefan-Boltzmann constant

e = Emissivity

T = Temperature (Kelvin)

Our values are given as

$A = 1.36m^2$

$\Delta T^4 = T_2^4 -T_1^4 = 307^4-T_1^4$

$\sigma = 5.67*10^{-8} J/(s m^2 K^4)$

$P = 122J/s$

$e = 0.7$

Replacing at our equation and solving to find the temperature 1 we have,

$P = \sigma Ae\Delta T^4$

$P = \sigma Ae (T_2^4 -T_1^4)$

$122 = (5.67*10^{-8})(1.36)(0.7)(307^4-T_1^4)$

$T_1 = 285.272K = 12.122\°C$

Therefore the the temperature of the coldest room in which this person could stand and not experience a drop in body temperature is 12°C

8 0

3 years ago

Which three things are related according to Newton's second law? force, mass, and acceleration force, matter, and acceleration f

Orlov [11]

Answer:

Force mass and acceleration

Explanation:

3 0

3 years ago

The critical angle for total internal reflection for cubic zirconia surrounded by air is 27.0°. Calculate the polarizing angle f

Anna11 [10]

Answer:

Polarizing angle, $\theta_{P} = tan^{- 1}{2.2} = 65.56^{\circ}$

Given:

Critical angle, $\theta_{cr} = 27^{\circ}$

Solution:

Now, in Total Internal Reflection (TIR), the critical angle for cubic zirconia is given by:

$sin\theta_{cr} = \frac{1}{\mu_{Z}}$ (1)

where

${\mu_{Z}$ = refractive index of zirconia

From eqn (1):

$\mu_{Z} = \frac{1}{sin\theta_{cr}}$

$\mu_{Z} = \frac{1}{sin(27^{\circ})} = 2.2$

Now, the angle of polarization is given by:

tan $\theta_{P} = \mu_{Z}$

Therefore,

$\theta_{P} = tan^{- 1}{2.2} = 65.56^{\circ}$

8 0

3 years ago

Brewed coffee is often too hot to drink right away. You can cool it with an ice cube, but this dilutes it. Or you can buy a devi

RUDIKE [14]

Answer:

75.2° C

Explanation:

Given:

mass of the coffee, M = 500

mass of the cup, m = 240

since the coffee is mostly water ,

specific heat of the coffee, C = 4.184 J/kg.°C

specific heat of the aluminium, C' = 0.902 J/kg.°C

The heat lost by the coffee will be transferred to the aluminium cup. The process of heat transfer will continue till the temperature of the coffee and the cup will be in equilibrium (or same)

i.e

Heat lost by the coffee = heat gained by the cup

MC(85 - T) = mC'(T - (-20))

where, T is the temperature at the equilibrium or the final temperature

on substituting the values, we get

500 × 4.184 × (85 - T) = 240 × 0.902 × (T - (-20))

T = 75.2° C

Hence, the final temperature of the coffee will be 75.2° C

6 0

3 years ago