A 20% efficient real engine is used to speed up a train from rest to 7.0 m/s. An ideal (Carnot) engine using the same cold and hot reservoir would accelerate the same train from rest to a speed of 8.0 m/s using the same amount of fuel. The cold reservoir is at 290 K. Find the temperature of the hot reservoir. Hint: Think work energy theorem.

To find the temperature of the hot reservoir, we can use the work-energy theorem and the efficiency of the engines.

First, let's start by calculating the work done by the real engine. The work done is equal to the change in kinetic energy. The formula is given by:

Work = (1/2) * m * v^2

where m is the mass of the train and v is the final velocity.

Given that the train is accelerated from rest to a speed of 7.0 m/s, and the mass of the train is not provided, we can assume it to be arbitrary, say 1 kg. So, we have:

Work = (1/2) * 1 kg * (7.0 m/s)^2
= 24.5 J

We know that the real engine is 20% efficient. The efficiency of an engine is given by the ratio of the useful work output to the energy input. So, we can write:

Efficiency = (Useful Work Output) / (Energy Input)

Since it's given that both the real engine and the ideal (Carnot) engine use the same amount of fuel, we can assume the energy input is the same. Therefore:

Efficiency of real engine = Efficiency of Carnot engine
0.20 = (Useful Work Output Carnot) / (Energy Input)

From this, we can conclude that the useful work output of the real engine is also 20% of the energy input.

Now let's consider the Carnot engine. The efficiency of a Carnot engine is given by:

Efficiency(Carnot) = 1 - (T(cold) / T(hot))

where T(cold) is the temperature of the cold reservoir and T(hot) is the temperature of the hot reservoir.

Plugging in the given values, we have:

0.20 = (Useful Work Output Carnot) / (Energy Input)
0.20 = (1 - (T(cold) / T(hot))) * (Energy Input / Energy Input)
0.20 = 1 - (T(cold) / T(hot))
T(cold) / T(hot) = 1 - 0.20
T(cold) / T(hot) = 0.80
T(cold) = 290 K

Now we can solve for T(hot):

0.80 = T(cold) / T(hot)
0.80 = 290 K / T(hot)
T(hot) = 290 K / 0.80
T(hot) ≈ 362.5 K

Therefore, the temperature of the hot reservoir (T(hot)) is approximately 362.5 K.