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ki77a [65]
3 years ago
6

A 1.0-kg block of aluminum is at a temperature of 50 Celsius. How much thermal energy will it lose when it’s temperature is redu

ced by half?
Physics
1 answer:
Ipatiy [6.2K]3 years ago
7 0

Answer:

The lose of thermal energy is, Q = 22500 J

Explanation:

Given data,

The mass of aluminium block, m = 1.0 kg

The initial temperature of block, T = 50° C

The final temperature of the block, T' = 25° C

The change in temperature, ΔT = 50° C - 25° C

                                                     = 25° C

The specific heat capacity of aluminium, c = 900  J/kg°C

The formula for thermal energy,

                             <em>Q = mcΔT</em>

                                 = 1.0 x 900 x 25

                                 = 22500 J

Hence, the lose of thermal energy is, Q = 22500 J

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Answer:

A. reintroducing an animal to the ecosystem

Explanation:

As generally, all know that for restoring an ecosystem naturally, it requires reintroduction of an animal to the ecosystem. As though it helps in reimposing the ecosystem back, and also helps to improve our ecosystem in natural surroundings, natural terrain, and population density. Basically reintroducing an animal is also required for the balancing of the ecosystem. As everything requires a properly balanced nature.

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] A new coal-fired 750 MWe power plant with a thermal efficiency of 42% burns 9000 Btu/lb coal, which contains 1.1% sulfur. a. I
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Answer:

The net emissions rate of sulfur is 1861 lb/hr

Explanation:

Given that:

The power or the power plant = 750 MWe

Since the power plant with a thermal efficiency of 42% (i.e. 0.42) burns 9000 Btu/lb coal, Then the energy released per one lb of the coal can be computed as:

\mathtt{=( 0.42\times 9000\times 1055.06) J}

= 3988126.8  J

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Also, The mass of the burned coal per sec can be calculated by dividing the molecular weight of the power plant by the energy released per one lb.

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The mass of the coal that is burned per sec =\dfrac{750}{3.99}

The mass of the coal that is burned per sec = 187.97 lb/s

The mass of sulfur burned  = \dfrac{1.1}{100} \times 187.97 \  lb/s

= 2.067 lb/s

To hour; we have:

= 7444 lb/hr

However, If a scrubber with 75% removal efficiency is utilized,

Then; the net emissions rate of sulfur is (1 - 0.75) × 7444 lb/hr

= 0.25 × 7444 lb/hr

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