A building is maintained at a temperature T by means of an ideal heat pump which uses a river at temperature T0 as a source of heat. The heat pump consumes power W, and the building loses heat to its surroundings at a rate α(T−T0), where α is a positive constant. Show that T is given by T=T0+W2α(1+1+4αT0/W‾‾‾‾‾‾‾‾‾‾‾√).

We want to prove that

Show that T is given by T=To+W/2α[1+√(1+4αTo/W)]

Explanation:

The efficiency of the pump is given as

n = Q/t / P

Where Q/t is the rejected power

And P is the consumed power to cool the house

Given that,

The at rejected or loss is α(T−To)

So, Q/t = α(T−To)

Pump Consumed =W

Then, power consumed P=W/t

Then,

n = Q/t /P

n = α(T−To) / W . Equation 1

Generally,

The efficiency of a carnot engine is given as

n = TH/(TH —TC)

Where TC is cold temperature

TC=To

And TH is high temperature

TH=T

n= T/ (T—To) Equation 2

Equation the two equation we have,

α(T−To) / W = T/ (T—To)

Cross multiply

α(T−To) × (T-To) =W× T

α(T² — 2TTo + To²)=WT

Divide both sides by α

T² — 2TTo + To²=WT/α

Rearrange

T² — 2TTo + To²—WT/α=0

T² — 2TTo —WT/α + To²=0

T² — (2To + W/α)T + To²=0

This form a quadratic equation

So let use formula method.

T = [- b ± √(b²-4ac) ]/ 2a

Where a=1, b= — (2To + W/α), c= To²

Then, applying the formula

T = [- b ± √(b²-4ac) ]/ 2a

T = [- -(2To + W/α) ± √(-(2To +W/α))² - 4×1×To²)]/ 2×1

T = [(2To + W/α) ± √4To²+4ToW/α+ W²/α² - 4To²]/ 2

T = [(2To+W/α) ±√(4ToWα+W²/α²)]/2

T=[(2To+W/α)±√(W²/α²(4Toα/W+1)]/2

T=[(2To+W/α)±W/α√(4Toα/W+1)]/2

Divide through by 2

T=To+W/2α±W/2α√(4Toα/W+1)

T=To+W/2α(1 ±√(4Toα/W+1)

Rearranging to conform to what we want to proof

T=To+W/2α(1 ±√(1 + 4αTo/W)

This is the required proof..

T = To + W/2α(1 + √(1 + 4αTo/W)

OR

T=To+W/2α(1 —√(1 + 4αTo/W)

T = (T_o + W/2α (1 + √(4αT_o/(W) + 1)] has been proved!

Explanation:

We know that efficiency of pump is given by;

η = (Q/t)/P

Where;

P = the power consumed in the house

Q/t = the rejected power.

Now, P can also be expressed as W/t

Likewise, Q/t can also be expressed as; α(T - T_o)

From the question, pump consumed power of W. Thus, P = W in this case.

So, efficiency can be expressed as;

η = (α(T - T_o))/(W) - - - - (1)

Now, for the carnot engine of the pump, efficiency is expressed as;

η = T/(T - T_o) - - - - (2)

Equating equation 1 and 2,we have;

(α(T - T_o))/(W) = T/(T - T_o)

α(T - T_o)² = WT

Expanding, we have;

T² - (2T•T_o) + T_o² = WT/α

T² - (2T•T_o + WT/α) + T_o² = 0

T² -T(2T_o + W/α) + T_o² = 0

To find T, let's use quadratic formula which is;

x = [-b ± √(b² - 4ac)]/2a

T = -(-(2T_o + W/α)) ± √(-(2T_o + W/α)² - (4•1•T_o²)]/(2•1)

T = (2T_o + W/α)) ± √((-2T_o - W/α)² - (4T_o²)]/(2)

T = (2T_o + W/α)) ± √((4T_o² + 2T_o•W/α + (W/α)² - 4T_o²)]/(2)

T = (2T_o + W/α)) ± √(2T_o•W/α + (W/α)²)]/(2)

T = (T_o + W/2α)) ± (1/2)√(2T_o•W/α + (W/α)²)]

T = (T_o + W/2α)) ± (W/2α)√(4T_o/(W/α) + 1)]

T = (T_o + W/2α)) ± (W/2α)√(4αT_o/(W) + 1)]

Now, let's arrange to correspond with what's in the question. And it means we will make use of the positive sign where we have ±.

Thus;

T = (T_o + W/2α (1 + 1√(4αT_o/(W) + 1)]

T = (T_o + W/2α (1 + √(4αT_o/(W) + 1)]

b. a lightning bug, plug of grass, and a worm