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umka21 [38]
3 years ago
8

Homes may be heated by pumping hot water through radiators. What mass of water (in g) will provide the same amount of heat when

cooled from 90.7 to 39.4°C, as the heat provided when 182 g of steam is cooled from 118°C to 100.°C. (Assume that the specific heat of liquid water is 4.187 J/g·°C over the given temperature range. Also assume that the specific heat of water vapor above its boiling point is 2.078 J/g·°C.)
Engineering
1 answer:
Nitella [24]3 years ago
3 0

Answer:

a mass of water required is mw= 1273.26 gr = 1.27376 Kg

Explanation:

Assuming that the steam also gives out latent heat, the heat provided should be same for cooling the hot water than cooling the steam and condense it completely:

Q = mw * cw * ΔTw = ms * cs * ΔTw + ms * L

where m = mass , c= specific heat , ΔT=temperature change, L = latent heat of condensation

therefore

mw = ( ms * cs * ΔTw + ms * L )/ (cw * ΔTw )

replacing values

mw = [182g * 2.078 J/g°C*(118°C-100°C) + 118 g * 2260 J/g ] /[4.187 J/g°C * (90.7°C-39.4°C)] = 1273.26 gr = 1.27376 Kg

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Answer:

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Explanation:

Solution

Code:

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public abstract class DiscountPolicy

{    

public abstract double computeDiscount(int count, double itemCost);

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at Override

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at Override

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Answer:

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Answer:

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int fd3[2];

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}

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 fprintf(stderr, "Pipe Failed" );

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