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Darasi game da electrical method

  • A makala ta da ta gabata mai suna method of mixture na yi bayanin specific heat capacity yanda ya ke dauke da method guda biyu na measuring dinshi wanda na kawo su kamar haka:

    1. Method of mixtures
    2. Electrical Method

    Saboda haka method din lissafinsu ma biyu ne, kuma na dauki guda daya nai bayaninsa shine wato method of mixture, ga kadan daga cikin bayaninshi kamar haka:

    Method of mixture shi ne a yayinda substance mai zafi da substance mai sanyi (solid/solid, solid/liquid, or liquid/liquid) suka hadu da junansu, the hotter substance loses heat to the colder substance. The transfer of heat  from the hotter to the colder substance  will continue until the two  substance attain the same temperature.

    Yau kuma zamu yi bayani ne akan electrical method sannan mu koyi yadda formula dinshi ya ke tare da yadda ake lissafinshi a karkashin ilmin kimiyyar lissafi wato physics. Kuma a karshe zamu amsa tambayoyi da suke fitowa a jarabawar karshe ta gama sakandire WAEC, NECO, JAMB kamar kullum.

    Electrical method

    Yana da kyau mai karatu ya duba text book na SS1 inda zaiyi revision akan abinda ya shafi electrical energy da kuma power, zai taimaka sosai da kuma amfani dangane da lissafin specific heat capacity tare da amfani da electrical method. A yayinda ake amfani da electrical heater wajan heating din substance  ko kuma substances (liquid, solid/liquid, or solid), heat energy (Q) wanda substance(s) din ya samu  is equal to the electrical energy (H) supplied by the heater. Saboda haka yanzu zamu kawo fomulolin na electrical energy muga yadda suke da kuma kowanne mene ne ma’anarsa a Turance.

    Heat gained by substance = heat supplied by electrical heater

    Q = H

    mc =IVt      or   mc (θ2 – θ1) = IVt

    mc =Pt        or     mc (θ2 – θ1) = Pt

    mc = I2Rt    or     mc (θ2 – θ1) = I2Rt

    mc = V2t / R   or  mc (θ2 – θ1) = V2t / R  

    where

    m = mass of substance in kilograms (kg)

    c = specific heat capacity in Jkg-1K-1  or Jkg-1

     = change in temperature in K or

    θ2 = final temperature after heating in K or

    θ1 = initial temperature before heating in K or

    I = current Amperes ( A )

    V = potential difference or voltage in volts ( V )

    t = duration of current flow in seconds (s)

    P = IV = electrical power ( power rating) in watt ( W )

    H = electrical heat energy in Joules (J)

    R = resistance of heating coil in ohm ( )

    Kowannen a cikin wadannan equation da na kawo za a iya amfani da shi wajan lissafi sai dai kawai a samu banbancin data wato abubuwan da aka bamu a cikin tambaya shi zai sa mugane wanne ya kamata muyi amfani da shi.

    Sannan a dubayadda ake measurement of heat capacity

    Yanzu zamu kawo misalai kai tsaye wanda suke fitowa a jarabawar karshe ta gama sakandire wato WAEC,NECO,JAMB wanda za su taimaka ma dalibi mai shirin gama sakandire dan samun cikakken fahimta danga ne da abubuwan da ake tambaya.

    Misali na daya:

    How long will it take to heat  3kg of water from 28 to 88 in an electric heater taking a current of 6A from an e.m.f source 220V? [ specific heat capacity of water is  given as = 4200Jkg-1K-1 ]. WAEC 1990

    Amsar tambaya:

    Yanzu zamu fara fitar da data kafin musan formula da zamuyi amfani da ita wajan amsa tambayar.

    Data:

    Mass of water, m = 3kg

    Current = 6A

    Initial temperature, 1 = 28

    c= 4200Jkg-1K-1

    voltage, V = 220V, 

    final temperature, 2 = 88

    heat energy supply by heater = heat gained by water

    IVt  = mc ( θ2 – θ1)

    Yanzu zamu dauko abubuwanda aka bamu kamar yadda muka fitar da su a data saimu sa a cikin wannan equation din.

    6 220  t = 3  4200 (88 – 28)

    1320  t = 12600  60

    t = 756000 / 1320 = 572.7s

    Misali na biyu:

    How long does it take a 750W heater operating at a full rating to raise the temperature of 1kg of water from 40 to 70 ? [ take the specific heat capacity of water as = 4200Jkg-1K-1  and neglect heat losses]. WAEC 1993

    Amsar Tambayar:

    Yanzu zamu fara fitar da abubuwan da aka bamu, wato data, hakan zai sa musan wane equation ne ya kamata mu yi amfani da shi.

    Data

    Mass of water, m = 1kg

    Power rating of heater, P = 750W;

    Initial temperature, 1 = 40;

    Final temperature, = 70 ;

    Specific heat capacity, c = 4200Jkg-1K-1

    Tunda mun samu fitar da data dinmu yanzu zamu sa equation dinmu sai mu saka abubuwan da aka bamu kowanne agunsa mu yi lissafi.

    Heat supplied by heater = heat gained by water

    Pt = mc (2 - 1 )

    750  t = 1 4200 ( 70 – 40)

    750  t = 4200  30

    t = 126000 / 750  = 168s

    Misali na uku:

    A heating coil of resistance 20Ω connected to a 220V source is used to boil a certain quantity of water in  a container of heat capacity 100JK-1  for 2 minutes. If the initial temperature of the water is 40 , calculate the mass of the water in the container. [specific heat capacity of water = 4.2  103  Jkg-1K-1 ; assume boiling point of water = 100].  WAEC 2002

    Amsar tambayar:

    Note; yana da kyau mu san cewa heat generated by the heater will be absorbed by the water and the container, don haka zai zama

    heat generated by heater = heat absorbed by water + heat absorbed by container

    Yanzu zamu fitar da abubuwan da aka bamu wato data

    Data:

    Resistance of heater , R = 20Ω

    Voltage, V = 220V

    Heat capacity of container, C = 100JK-1  

    Time, t = 2min = 2  60s = 120s

    Final tempt, θ2 = 100 (boiling point of water).

    Initial tempt., θ1 = 40;

    Specific heat capacity of water, c = 4200Jkg-1K-1

    Yanzu zamu dauki equation din da ya kamata muyi amfani dashi,

    Let m be the mass of water

    Substitute into V2t / R = [mc (2 - 1 ) ]water  +  [  (2 - 1 ) ]container

    Remember: quantity of heat (Q) = heat capacity ()  temperature change ( ) 

    Yanzu zamu dauko data dinmu kowanne abun da aka bamu musa shi cikin equation dinmu sai muyi lissafinsu.

    2202 /20    120 = m 4200 (100 – 40 ) + 100 ( 100 – 40 )

    290400 = m  4200  60 + 100  60

    290400 = m  252000 + 6000

    290400 – 6000 = 252000m

    284400 = 252000m

    m = 284400 / 252000 = 1.13kg

    Za a iya duba wannan makala da ta yi takaitaccen bayani akan gravitational field

    Wadannan kadanne daga cikin misalan lissafin electrical method yana da kyau dalibai su dauko past questions na physics su duba wanda suke magana akan electrical method domin suyi practicing amsa tambayoyin da daban-daban. Ku kasance tare da ni ako da yaushe don kara fahimta da kuma koyon ilmin kimiyyar lissafi wato physics, Allah kara mana ilmi mai amfani ameen.

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