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gtnhenbr [62]
3 years ago
15

The air in an inflated balloon (defined as the system) is warmed over a toaster and absorbs 130 J of heat. As it expands, it doe

s 79 kJ of work. What is the change in internal energy for the system? Express the energy in kilojoules to two significant figures.
Physics
1 answer:
Nadya [2.5K]3 years ago
8 0

Answer:

The change in internal energy for the system = - 78.87 KJ = - 79 KJ to 2 s.f

Explanation:

The first law of thermodynamics states that energy can neither be created nor destroyed but can only be transformed from one form to another.

It goes further to explain that change in the internal energy of a system (ΔU) is the sum of heat exchanged (Q) and the work done (W) by the system or on the system.

If work is done by the system, sign of work is ‘-’, if work is done on the system, the sign of work is ‘+’.

If heat is released by the system, the sign of q is ‘-’. And if heat is gained by the system, then the sign of q is ‘+’.

ΔU = Q + W

Q = 130 J

W = - 79 KJ = - 79000 J

ΔU = 130 - 79000 = - 78870 J = - 78.87 KJ = - 79 KJ

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The electronics supply company where you work has two different resistors, R1 and R2, in its inventory, and you must measure the
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Answer:

R₁ = 23.77 ohms and R₂ = 7.92 ohms

Explanation:

When connected in series, the equivalence resistance, R(eq) is given as

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When connected in parallel, the equivalence resistance, R(eq) is given as

[1/R(eq)] = [(1/R₁) + (1/R₂)]

R(eq) = (R₁R₂)/(R₁ + R₂)

The parallel and series combination are connected to a battery of emf 39.0 V with negligible internal resistance. And the power supplied is measured.

But power supplied is given as

P = IV = (V/R) V = (V²/R)

When connected in series, the power supplied is given as

P = 48.0 W,

V = 39.0 V,

R = R(eq) = (R₁ + R₂)

48 = (39²/R)

R = (39²/48)

R = 31.6875 ohms

R = (R₁ + R₂) = 31.6875

(R₁ + R₂) = 31.6875 (eqn 1)

When connected in series, the power supplied is given as

P = 256.0 W,

V = 39.0 V,

R = R(eq) = (R₁R₂)/(R₁ + R₂)

256 = (39²/R)

R = (39²/256)

R = 5.9414 ohms

R = R(eq) = (R₁R₂)/(R₁ + R₂) = 5.9414

(R₁R₂)/(R₁ + R₂) = 5.9414

But, recall eqn 1

(R₁ + R₂) = 31.6875

(R₁R₂)/(R₁ + R₂) = 5.9414

Substituting for (R₁ + R₂)

(R₁R₂)/(R₁ + R₂) = (R₁R₂)/31.6875 = 5.9414

(R₁R₂) = 31.6875 × 5.9414 = 188.2683

R₁ = (188.2683/R₂)

(R₁ + R₂) = 31.6875

Substituting for R₁

(188.2683/R₂) + R₂ = 31.6875

multiply through by R₂

188.2683 + R₂² = 31.6875R₂

R₂² - 31.6875R₂ + 188.2683 = 0

Solving the quadratic equation

R₂ = 23.77 ohms or 7.92 ohms

If R₂ = 23.77 ohms, R₁ = 7.92 ohms

If R₂ = 7.92 ohms, R₁ = 23.77 ohms

Since the question explains that R₁ > R₂

R₁ = 23.77 ohms and R₂ = 7.92 ohms

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