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Verizon [17]
3 years ago
10

Calculate the rate at which body heat is conducted through the clothing of a skier in a steady- state process, given the followi

ng data: the body surface area is 1.80 m and the clothing is 1.00 cm thick; the skin surface temperature is 33.0 C and the outer surface of the clothing is at 1.00 C the thermal conductivity of the clothing is 0.040 W/m K
Engineering
1 answer:
olga2289 [7]3 years ago
4 0

Answer:

230.4W

Explanation:

Heat transfer by conduction consists of the transport of energy in the form of heat through solids, in this case a jacket.

the equation is as follows

Q=\frac{KA(T2-T1)}{L} \\

Where

Q=heat

k=conductivity=0.04

A=Area=1.8m^2

T2=33C

T1=1C

L=thickness=1cm=0.01mQ=\frac{(0.04)(1.8m^2)(33-1)}{0.01m}

Q=230.4W

the skier loses heat at the rate of 230.4W

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The future perfect tense forms are made by putting ‘will / shall + have’ before the past participle from the verb. these sentences can be changed into the passive if the active verb has an object


i hope this helps :D thanks
5 0
3 years ago
Describe, in a general form, the equation, in time domain, that tells the voltage across a inductor, L, as a function of time wh
love history [14]

Answer:

a) V(t) = Ldi(t)/dt

b) If current is constant, V = 0

Explanation:

a) The voltage, V(t), across an inductor is proportional to the rate of change of the current flowing across it with time.

If  V represents the Voltage across the inductor

and i(t) represents the current across the inductor in time, t.

V(t) ∝ di(t)/dt

Introducing a proportionality constant,L, which is the inductance of the inductor

The general equation describing the voltage across the inductor of inductance, L, as a function of time when a current flows through it is shown below.

V(t) = Ldi(t)/dt ..................................................(1)

b) If the current flowing through the inductor is constant i.e. does not vary with time

di(t)/dt = 0   and hence the general equation (1) above becomes

V(t) = 0

4 0
3 years ago
An aluminum metal rod is heated to 300oC and, upon equilibration at this temperature, it features a diameter of 25 mm. If a tens
Natalka [10]

Answer:

It will results in mechanical hardening.

5 0
3 years ago
Read 2 more answers
14. The top plate of the bearing partition
aliina [53]

Answer:

d. is applied after the ceiling joists are

installed.

7 0
3 years ago
A large heat pump should upgrade 5 MW of heat at 85°C to be delivered as heat at 150°C. Suppose the actual heat pump has a COP o
AysviL [449]

Answer:

W=2 MW

Explanation:

Given that

COP= 2.5

Heat extracted from 85°C  

Qa= 5 MW

Lets heat supplied at 150°C   = Qr

The power input to heat pump = W

From first law of thermodynamics

Qr= Qa+ W

We know that COP of heat pump given as

COP=\dfrac{Qr}{W}

2.5=\dfrac{5}{W}

2.5=\dfrac{5}{W}

W=2 MW

For Carnot heat pump

COP=\dfrac{T_2}{T_2-T_1}

2.5=\dfrac{T_2}{T_2-(273+85)}

2.5 T₂ -  895= T₂

T₂=596.66 K

T₂=323.6 °C

7 0
3 years ago
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