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Heat and TheromdynamicsQuestion and Answers: Page 1

Question Number 214780    Answers: 0   Comments: 3

Hello, friends rm, in fact, just a day old here. I need p into MS-Word H.C.V = (((m_w + m_e ) c_w (t_2 − t_1 ))/m_f ). = (((2.5 + 0.39) 4.2 × 2.6)/(0.001)) kJ/kg = (((2.89 ) × (4.2 × 2.6))/(0.001)) kJ/kg = ((31.5588)/(0.001)) kJ/kg = 31 558.8 kJ/kg How do I transfer this typed formula and calculation into microsoft word format? Thanks!

$$\mathrm{Hello},\:\mathrm{friends} \mathrm{rm},\: \\ $$$$\mathrm{in}\:\mathrm{fact},\:\mathrm{just}\:\mathrm{a}\:\mathrm{day}\:\mathrm{old}\:\mathrm{here}. \\ $$$$\mathrm{I}\:\mathrm{need} \\ $$$$ \mathrm{p} \:\mathrm{into}\:\mathrm{MS}-\mathrm{Word}\: \\ $$$$ \\ $$$$ \\ $$$$\:\:\mathrm{H}.\mathrm{C}.\mathrm{V}\:=\:\frac{\left(\mathrm{m}_{\mathrm{w}} \:+\:\mathrm{m}_{\mathrm{e}} \right)\:\mathrm{c}_{\mathrm{w}} \:\left(\mathrm{t}_{\mathrm{2}} \:−\:\mathrm{t}_{\mathrm{1}} \right)}{\mathrm{m}_{\mathrm{f}} }. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\left(\mathrm{2}.\mathrm{5}\:+\:\mathrm{0}.\mathrm{39}\right)\:\mathrm{4}.\mathrm{2}\:×\:\mathrm{2}.\mathrm{6}}{\mathrm{0}.\mathrm{001}}\:\mathrm{kJ}/\mathrm{kg} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\left(\mathrm{2}.\mathrm{89}\:\right)\:×\:\left(\mathrm{4}.\mathrm{2}\:×\:\mathrm{2}.\mathrm{6}\right)}{\mathrm{0}.\mathrm{001}}\:\mathrm{kJ}/\mathrm{kg} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{31}.\mathrm{5588}}{\mathrm{0}.\mathrm{001}}\:\mathrm{kJ}/\mathrm{kg} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\underline{\mathrm{31}\:\mathrm{558}.\mathrm{8}\:\boldsymbol{\mathrm{kJ}}/\boldsymbol{\mathrm{kg}}} \\ $$$$ \\ $$$$\mathrm{How}\:\mathrm{do}\:\mathrm{I}\:\mathrm{transfer}\:\mathrm{this}\:\mathrm{typed}\:\mathrm{formula}\:\mathrm{and}\: \\ $$$$\mathrm{calculation}\:\mathrm{into}\:\mathrm{microsoft}\:\mathrm{word}\:\mathrm{format}?\: \\ $$$$\mathrm{Thanks}! \\ $$

Question Number 206171    Answers: 0   Comments: 0

Question Number 206168    Answers: 0   Comments: 1

Question Number 200129    Answers: 0   Comments: 0

Question Number 194194    Answers: 0   Comments: 0

Question Number 186028    Answers: 0   Comments: 2

A wolf weighs 40 kg, how many kilo calories does it need to maintain its body temperature?

$$ \\ $$A wolf weighs 40 kg, how many kilo calories does it need to maintain its body temperature?

Question Number 178587    Answers: 0   Comments: 0

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Question Number 164137    Answers: 0   Comments: 1

what is the triple point of water?

$${what}\:{is}\:{the}\:{triple}\:{point}\:{of}\:{water}? \\ $$

Question Number 141111    Answers: 1   Comments: 1

A gap must be left between steel rails to allow for thermal expansion. How large a gap is needed if the maximum temperature reached is 50° above the temperature at which the rails were laid. The length of the rail is 10m and the α_(steal) =12×10^(−6) /°c

$$\mathrm{A}\:\mathrm{gap}\:\mathrm{must}\:\mathrm{be}\:\mathrm{left}\:\mathrm{between}\:\mathrm{steel}\: \\ $$$$\mathrm{rails}\:\mathrm{to}\:\mathrm{allow}\:\mathrm{for}\:\mathrm{thermal}\:\mathrm{expansion}. \\ $$$$\:\mathrm{How}\:\mathrm{large}\:\mathrm{a}\:\mathrm{gap}\:\mathrm{is}\:\mathrm{needed}\:\mathrm{if}\:\mathrm{the}\: \\ $$$$\mathrm{maximum}\:\mathrm{temperature}\:\mathrm{reached} \\ $$$$\:\mathrm{is}\:\mathrm{50}°\:\mathrm{above}\:\mathrm{the}\:\mathrm{temperature}\:\mathrm{at}\: \\ $$$$\mathrm{which}\:\mathrm{the}\:\mathrm{rails}\:\mathrm{were}\:\mathrm{laid}.\:\mathrm{The}\: \\ $$$$\mathrm{length}\:\mathrm{of}\:\mathrm{the}\:\mathrm{rail}\:\mathrm{is}\:\mathrm{10m}\:\mathrm{and} \\ $$$$\:\mathrm{the}\:\alpha_{\mathrm{steal}} =\mathrm{12}×\mathrm{10}^{−\mathrm{6}} /°\mathrm{c} \\ $$

Question Number 137317    Answers: 0   Comments: 3

Question Number 136487    Answers: 0   Comments: 1

Question Number 131672    Answers: 0   Comments: 0

Δt=210C^° k=?

$$\Delta{t}=\mathrm{210}{C}^{°} \:\:\:\:\:\:\:\:\:{k}=? \\ $$

Question Number 127423    Answers: 0   Comments: 0

Question Number 127336    Answers: 0   Comments: 0

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Question Number 119452    Answers: 1   Comments: 1

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Question Number 103409    Answers: 0   Comments: 1

What will be the state of a system(isolated)when its entropy tends to zero or zero? Time get stopped?? or another things may happen I am thinking about this

$$\mathrm{What}\:\mathrm{will}\:\mathrm{be}\:\mathrm{the}\:\mathrm{state}\:\mathrm{of}\:\mathrm{a}\:\mathrm{system}\left(\mathrm{isolated}\right)\mathrm{when}\:\mathrm{its}\:\mathrm{entropy}\:\mathrm{tends} \\ $$$$\mathrm{to}\:\mathrm{zero}\:\mathrm{or}\:\mathrm{zero}? \\ $$$$ \\ $$$$\mathrm{T}\boldsymbol{\mathrm{ime}}\:\boldsymbol{\mathrm{get}}\:\boldsymbol{\mathrm{stopped}}?? \\ $$$$\boldsymbol{\mathrm{or}}\:\boldsymbol{\mathrm{another}}\:\boldsymbol{\mathrm{things}}\:\boldsymbol{\mathrm{may}}\:\boldsymbol{\mathrm{happen}} \\ $$$$\boldsymbol{\mathrm{I}}\:\boldsymbol{\mathrm{am}}\:\boldsymbol{\mathrm{thinking}}\:\boldsymbol{\mathrm{about}}\:\boldsymbol{\mathrm{this}} \\ $$

Question Number 94922    Answers: 1   Comments: 0

Question Number 90073    Answers: 0   Comments: 0

$$ \\ $$

Question Number 75648    Answers: 0   Comments: 0

A student seals 200g of ice−cold water in a glass vacuum (thermos) flask and finds that it warms up by 3.5 K in one hour SHC of water =2400 Jkg^(−1) K^(−1) a) Calculate the average rate of heat flow in the flask in watts. To check this result over a longer period,the student fills the flask with equal amounts of ice and water all at 0°C, and leaves it for 4 hours. the specific latent heat(enthalpy) of fusion of ice is 0.33MJg^(−1) b) How much ice would you expect to have melted at the end of the four hours?

$$\:{A}\:{student}\:{seals}\:\mathrm{200}{g}\:{of}\:{ice}−{cold}\:{water}\:{in}\:{a}\:{glass}\:{vacuum} \\ $$$$\left({thermos}\right)\:{flask}\:{and}\:{finds}\:{that}\:{it}\:{warms}\:{up}\:{by}\:\mathrm{3}.\mathrm{5}\:{K}\:{in}\:{one}\:{hour} \\ $$$${SHC}\:{of}\:{water}\:=\mathrm{2400}\:{Jkg}^{−\mathrm{1}} {K}^{−\mathrm{1}} \\ $$$$\left.{a}\right)\:{Calculate}\:{the}\:{average}\:{rate}\:{of}\:{heat}\:{flow}\:{in}\:{the}\:{flask}\:{in}\:{watts}. \\ $$$${To}\:{check}\:{this}\:{result}\:{over}\:{a}\:{longer}\:{period},{the}\:{student}\:{fills}\:{the}\:{flask}\:{with} \\ $$$${equal}\:{amounts}\:{of}\:{ice}\:{and}\:{water}\:{all}\:{at}\:\mathrm{0}°{C},\:{and}\:{leaves}\:{it}\:{for}\: \\ $$$$\mathrm{4}\:{hours}.\:{the}\:{specific}\:{latent}\:{heat}\left({enthalpy}\right)\:{of}\:{fusion}\:{of}\:{ice}\:{is}\:\mathrm{0}.\mathrm{33}{MJg}^{−\mathrm{1}} \\ $$$$\left.{b}\right)\:{How}\:{much}\:{ice}\:{would}\:{you}\:{expect}\:{to}\:{have}\:{melted}\:{at}\:{the}\:{end}\:{of}\:{the} \\ $$$${four}\:{hours}? \\ $$

Question Number 75204    Answers: 0   Comments: 0

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