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True [87]
3 years ago
10

Carnot engine A has an efficiency of 0.57, and Carnot engine B has an efficiency of 0.72. Both engines utilize the same hot rese

rvoir, which has a temperature of 615 K and delivers 1248 J of heat to each engine. Find (a), (b) the magnitude of the work produced by engine A and B respectively and (c), (d) the temperatures of the cold reservoirs that they use.
Physics
1 answer:
dem82 [27]3 years ago
4 0

Answer:

710.79 Joules

898.56 Joules

264.45 K

172.2 K

Explanation:

W = Work done

Q = Heat added = 1248 J

T_h = Temperature of hot reservoir = 615 K

T_c = Temperature of cold reservoir

Efficiencies of engine A and B

\eta_A = 0.57

\eta_B = 0.72

Efficiency is given by

\eta=\frac{W}{Q}\\\Rightarrow W=\eta Q\\\Rightarrow W=0.57\times 1248\\\Rightarrow W=711.36\ J

Work done by engine A is 710.79 Joules

\eta=\frac{W}{Q}\\\Rightarrow W=\eta Q\\\Rightarrow W=0.72\times 1248\\\Rightarrow W=898.56\ J

Work done by engine B is 898.56 Joules

Efficiency

\eta=1-\frac{T_c}{T_h}\\\Rightarrow 0.57=1-\frac{T_c}{615}\\\Rightarrow T_c=-0.43\times -615\\\Rightarrow T_c=264.45\ K

The temperature of the cold reservoir is 264.45 K

\eta=1-\frac{T_c}{T_h}\\\Rightarrow 0.72=1-\frac{T_c}{615}\\\Rightarrow T_c=-0.28\times -615\\\Rightarrow T_c=172.2\ K

The temperature of the cold reservoir is 172.2 K

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A 6.0-kilogram block, sliding to the east across a horizontal, frictionless surface with a momentum of 30.0 kilogram · meters pe
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The final speed of the block after the collision with the obstacle is \boxed{3.33\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}}.

Further Explanation:

Given:

The mass of the block is 6.0\,{\text{kg}}.

The initial momentum of the block is 30\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/ {\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}.

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Concept:

The block is sliding towards east and the impulse imparted by the obstacle is towards the obstacle is towards west on the block. It means that the impulse exerted by the obstacle will reduce the momentum of the block.

According to the impulse momentum theorem, the rate of change of momentum of the body is equal to the impulse imparted to the body.

The expression for the impulse momentum theorem is.

{p_f} - p{ & _i} = I               …… (1)                                    

Substitute 30\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/{\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}} for {p_i} and - 10\,{\text{N}} \cdot {\text{s}} for I  in equation (1).

 \begin{aligned}{p_f} &= - 10\,{\text{N}} \cdot {\text{s}} + 30\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/{\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}} \\&= 20\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/{\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}\\\end{aligned}

The final momentum of the block can be expressed as:

{p_f} = m{v_f}                   …… (2)                                  

Substitute 20\text{kg}\;\text{m/s} for {p_f} and 6.0\,{\text{kg}} for m in equation (2).

 \begin{aligned}20 &= 6 \times {v_f} \\ {v_f}&= \frac{{20}}{6}\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}\\&= 3.33\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}} \\ \end{aligned}

Thus, the final speed of the block after the collision with the obstacle is \boxed{3.33\;\text{m/s}}.

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Answer Details:

Grade: High School

Chapter: Impulse-momentum theorem

Subject: Physics

Keywords:  Impulse, imparted, obstacle, speed, momentum, the obstacle, impulse-momentum theorem, frictionless surface, speed of block after collision.

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