Question

Think! Using the conversion rate of 4.19 joules per calorie, how much energy is necessary to heat 2.50 liters of water in a steam engine from 100°C to steam at 250°C?

Answer 1

Answer on Question # 74193, Physics -Mechanics- Relativity:

Question: Think! Using the conversion rate of 4.19 joules per calorie, how much energy is necessary to heat 2.50 litres of water in a steam engine from $100{}^{\circ}\mathrm{C}$ to steam at $250{}^{\circ}\mathrm{C}$?

Solution: Let, $Q_{1} =$ heat required to vaporize the water to steam at $100{}^{\circ}\mathrm{C}$

$Q_{2} =$ heat required to warm the steam from $100{}^{\circ}\mathrm{C}$ to $250{}^{\circ}\mathrm{C}$

$D =$ density of water $= 1000\mathrm{Kg / m^3}$

$V =$ volume of water 2.5 litres

$L_{v} =$ latent heat of vaporisation of water 2264.705 KJ/Kg

$s =$ specific heat of steam $= 4.19\mathrm{KJ / (Kg\cdot{}^{\circ}C)}$

$\Delta T =$ temperature difference $= 250{}^{\circ}\mathrm{C} - 100{}^{\circ}\mathrm{C} = 150{}^{\circ}\mathrm{C}$

So, mass of water $(\mathrm{m}) = \mathrm{V} \times \mathrm{D} = 2.5$ litres $\times 1000 \mathrm{~Kg} / \mathrm{m}^{3} = 2500$ litres . Kg/1000 litres $= 2.5 \mathrm{~Kg}$ (1 m³ = 1000 litres)

Now, $Q_{1} = m \times L_{v} = 2.5 \times 2264.705 = 5661.76$ KJ

And $Q_{2} = m \times s \times \Delta T = 2.5 \times 4.19 \times 150 = 1571.25$ KJ

Total energy $= Q_{1} + Q_{2} = 5661.76 + 1571.25 = 7233.01$ KJ

Answer: 7233.01 KJ energy is required.

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