Physics Question. Latent Heat and mixtures (warning: This is fairly hard stuff)?

Ok, can som1 show me(with full working) how to answer this questions, I have a physics exam soon and I have completly forgotten how to solve these problems, I will vote on the best person on two factors, it looks correct and detail on what they have done (setting out also helps)! Thankyou, The Question is as follows:



';You find that you have let a 12.0kg stainless steel barbique plate become too hot for normal cooking. You Decide to cool the plate from 296Deg C to 185Deg C by spraying water onto the plate. Specific heat of steel = 445 j/Kg K. Calculate the mass of water at 20Deg C yo will need, asuming all the water evaporate to steam at 100Deg C.



I thankyou for your time and I hope somone can answer me, please no answers that are not either solutions or links to direct how to's for this type of question (not a broad physics website).Physics Question. Latent Heat and mixtures (warning: This is fairly hard stuff)?
Im no physics Expert, but shouldn't have you included the energy used after water vaporisation?



let (m) kg of water spray (at 20 C) does the job of cooling the plate to desired temp%26gt; assuming all such water evaporated at 100 C

------------------------------------

energy conservation gives%26gt;.

sensible heat lost by plate = heat gained by water in raising its own temp + getting transferred from liquid to vapour phase



ms * cs * [296 - 185] = m * cw * [100 - 20] + m L (WHAT ABOUT AFTER VAPORISATION!!!!!!!!!!!!!!!!!

where ms = 12 kg, cs = 445 SI units, L = water = 336*1000 SI units, and cw = 4182 SI units



(you may like to use book data of water)



12*445[111] = m [4182*80 + 336000]

m = 592740/670560

m = 0.8839 kg

m = 883.95 gram waterPhysics Question. Latent Heat and mixtures (warning: This is fairly hard stuff)?
The net change in energy must = 0, that is:



The energy lost by the steel (negative value) + the energy gained by the water as it heats to 100 degrees + the energy gained by the water as it evaporates = 0



The energy lost/gained without a phase change =

specific heat x mass x change in temperature

The energy lost/gained through a phase change =

latent heat x mass.



Let m = mass of water (Specific heat = 4180 j/kg K and latent heat of vaporization = 2.26 x 10^6 j/kg





(445)(12)(185-296) + 4180) + 4180(m)(100-20) + 2.26E6(m) = 0



Solve for m.
Specific capacity of water = 4200 J / kg / degree celcius

Specific latent heat of vaporisation of water = 226 x 10^4 J / kg



The energy loss from steel = mc(tethre), (tethre) is the change in temperature.



Energy loss = (12)(445)[(296 + 273K) - (185 + 273K)] = 592 740 J



Energy gain by water from 20 deg C to 100 deg C = mc(tethre)



Energy gain 20degC to 100degC = (m)(4200)(100 - 20) J



Energy gain when water change to steam = mL, L is the latent heat of vaporization.



Energy gain when changes to steam = (m)(226 x 10^4 )



Energy loss from steel = Total energy gain by water

592 740 = (m)(4200)(100 - 20) + (m)(226 x 10^4 )

592 740 = 336000m + 2260000m

m = 592 740 / (336000 + 2260000) = 0.228 kg





Hope this is correct ^^
let (m) kg of water spray (at 20 C) does the job of cooling the plate to desired temp%26gt; assuming all such water evaporated at 100 C

------------------------------------

energy conservation gives%26gt;.

sensible heat lost by plate = heat gained by water in raising its own temp + getting transferred from liquid to vapour phase



ms * cs * [296 - 185] = m * cw * [100 - 20] + m L

where ms = 12 kg, cs = 445 SI units, L = water = 2270*1000 SI units, and cw = 4182 SI units



(you may like to use book data of water)



12*445[111] = m [4182*80 + 2270000]

m = 592740/2604560 (kg)

in grams

m = 227.58 gram water

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