1.

A geyser heats water flowing at the rate of `3.0` litre per minute from `27^(@)C` to `77^(@)C`. If the geyser operates on a gas burner and its heat of combustion is `4.0 xx 10^(4) J//g`, then what is the rate of combusion of fuel (approx.)?

Answer» Given, Volume of water heated,
V`=3.0 Lmin^(-1)` = `3 xx10^(-3)m^(3)min^(-1)`
As Density, `rho` = `(Mass(m))/(Volume(V))`
`therefore` Mass of wate heated,
`m=rhoV = 10^(3)xx3xx10^(-3)kg min^(-1)`
Rise in temperature of water,
`DeltaT = (T_(2)-T_(1)) = 77-21 = 50^(@)C`
Specific heat of water, `c = 4.2Jg^(-1)C^(-1)` = `4.2xx10^(3)Jkg^(-2)C^(-1)`
Heat required by wate, `DeltaQ = mcDeltaT`
`=3xx4.2xx106(3)xx50`Jmin
`=63 xx10^(4)Jmin^(-1)`................(i)
Heat of combustion of fuel = `4.0 xx 10^(4)Jg^(-1)` = `4.0 xx 10^(7)Jkg^(-1)`
Let m kg `min^(-1)` be the rate of combustion of fuedl
`= mxx4 xx10^(7Jmin^(-1)`...............(ii)
`therefore` Eqs. (i) and (ii) , we get
`63 xx10^(4)=mxx4xx10^(7)`
or `m=(63xx10^(4))/(4 xx10^(7)` = `15.75 xx 10^(-3)kg min^(-1)`
or rate of combution of fuel = `16g min^(-1)`


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