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blondinia [14]
2 years ago
10

Which client body temperatures are indicative of moderate hypothermia? select all that apply. one, some, or all responses may be

correct.
Physics
1 answer:
Lyrx [107]2 years ago
3 0

The correct option is (3)  88° F (31.1° C) and (4) 92° F (33.3° C).

A medical emergency known as hypothermia happens when your body results in a dangerously low body temperature.

What is Hypothermia?

  • After being exposed to chilly, rainy, or windy environments, hypothermia develops.
  • Your body eventually exhausts its reserves of energy after prolonged exposure to cold temperatures, which causes your body temperature to drop.
  • When the body's temperature falls below 95° F (35° C), hypothermia sets in. The average body temperature is 37° C (98.6° F).
  • A medical emergency is hypothermia. The brain and body cannot operate normally when a person's body temperature is dangerously low. Hypothermia can cause cardiac arrest (heart stops beating) and death if it is not addressed.

Learn more about the Hypothermia with the help of the given link:

brainly.com/question/13023361

#SPJ4

I understand that the question you are looking for is "Which client body temperatures are indicative of moderate hypothermia? Select all that apply.

1. 80° F (26.7° C)

2. 84° F (28.9° C)

3. 88° F (31.1° C)

4. 92° F (33.3° C)

5. 96° F (35.6° C)"

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A storage tank holds methane at 120 K, with a quality of 25 %, and it warms up by 5°C per hour due to a failure in the refrigera
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One of the fundamental pillars to solve this problem is the use of thermodynamic tables to be able to find the values of the specific volume of saturated liquid and evaporation. We will be guided by the table B.7.1 'Saturated Methane' from which we will obtain the properties of this gas at the given temperature. Later considering the isobaric process we will calculate with that volume the properties in state two. Finally we will calculate the times through the differences of the temperatures and reasons of change of heat.

Table B.7.1: Saturated Methane

T_1 = 120K

p_1 = 191.6kPa

v_f = 0.002439m^3/kg

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v_1 = v_f+x_1v_{fg}

v_1 = 0.002439+ (0.25)(0.30367)

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Assume the tank is rigid, specific volume remains constant

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v_2 = 0.0783m^3/kg

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p_2 = 823.7kPa

We can calculate the time taken for the methane to become a single phase

t = \frac{T_2-T_1}{\dot{T}}

Here

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\dot{T} = Warming rate

Replacing

t = \frac{(145-273)-(120-273)}{5}

t = \frac{25}{5}

t = 5hr

Therefore the time taken for the methane to become a single phase is 5hr

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