## Darasi game da method of mixtures

• A makala ta da ta gabata mai suna measurement of heat capacity na yi bayani dangane da specific heat capacity of a body a matsayin, Specific Heat Capacity of a body. Idan quantity na heat energy, Q joules, yasa temperature na body mai dauke da mass M kg ya yi rising daga θ10C zuwa θ2, ana samu cewa Q is proportional to M sannan temperature yana canzawa (θ21), sannan kuma yana depending ne akan nature of the body. A turance ana defining din specific heat capacity ne kamar haka:

The specific heat capacity of a substance is defined as the quantity of heat required to raise the temperature of 1 kg of the substance through one degree centigrade or one Kelvin.for water, the specific heat capacity is 4200 J/kg or 4.2 J/gk. The specific heat capacity of water is higher than that of many other substances.

A yau kuma zamu ci gaba da bayani dangane da specific heat capacity inda ya kamata mu sani cewa ana measuring specific heat capacity ne da method guda biyu, ga su kamar haka;

1. Method of mixtures
2. Electrical Method

Saboda haka method din lissafin su ma biyu ne yanzu zamu fara daukar guda dayansu muga yadda ake lissafin shi.

Method of mixtures

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. Therefore, we can state the following;

Heat given out by hotter substance = Heat gained by colder substance

that is, QH = QC

let H be the temperature of the hotter substance

θC be the temperature of the colder substance

θ be the final or equilibrium temperature of both substances or mixture.

The temperature change of the hotter substance is

θH = θH -  θ

The temperature change of the colder substance is

θC = θ – θC

Heat given out by hotter substance, QH = mcθH

QH = mc (θH - &theta Heat gained by colder substance, QC = mcθC

QC = mc (θ – θC )

If heat given out = Heat gained

Then QH = QC

mcθH = mcθC

mc (θH - &theta = mc (θ – θC )

The masses and specific heat capacities of the hotter and colder substances could be different of the same.

Misali na daya:

Calculate the final or equilibrium temperature of the mixture of 250g of water 65 is added to 150g of colder water at 5. Neglect heat absorbed by the surrounding. [ specific heat capacity of water = 42000 Jkg-1K-1 .

Amsar Tambaya:

Data:

Mass of hot water, m = 250g = 0.25kg,             Mass of cold water,m = 150g = 0.15kg

Temperature of hot water, θH = 65  ,                  Temperature of cold water, θC = 5

SHC of water, c = 4200 Jkg-1K-1,                     SHC of water , c = 4200 Jkg-1K-1

Yanzu zamu dauki θ ya zama a matsayin final or equilibrium temperature.

Heat given out by hot water = heat gained by cold water

QH = QC

mc (θH - &theta = mc (θ – θC)

0.25  4200  (65 – &theta = 0.15  4200  (θ – 5)

Divide both sides by 4200 to obtain

0.25 (65 – ) = 0.15 (  - 5 )

16.25 – 0.25 = 0.15θ – 0.75

16.25 + 0.75 = 0.15θ + 0.25θ

17.0 = 0.4θ

θ = 17.0 / 0.4 = 42.5

Misali na biyu:

Hot water is added to four times its mass of water at 25 and thoroughly stirred. If the final temperature of the mixture is 40 , calculate the initial temperature of the hot water.

Amsar Tambaya:

Data:

Yanzu zamu fara fitar da abubuwan da aka bamu a cikin tambayar

Mass of hot water = m                                    Mass of cold water = 4m

Temperature of hot water, θH                         Temperature of cold water, θC = 25

SHC of hot water = c                                      SHC of cold water = c

Final temperature of mixture, θ = 40

Heat given out by hot water = heat gained by cold water

QH = QC

mc (θH - &theta = mc (θ – θC )

mc (θH - 40) = 4mc (40 – 25)

divide both sides by mc to obtain

θH - 40 = 4 × 15

θH - 40 = 60

θH = 60 + 40 = 100

Misali na uku:

A tap supplies water at 15 while another supplies water at 90. If a man wishes to bathe with water at 30, calculate the ratio of the mass of cold water to the mass of hot water.

Amsar tambaya:

Yanzu zamu fara fitar da abubuwan da aka bamu a cikin tamyar

Data:

Mass of water = mH ,                                 mass of cold water = mC

Temperature of hot water, θH = 90        Temperature of cold water, θC = 15

SHC of water = c                                      SHC of water = c

Final temperature of mixture of cold / hot water, θ = 30.

Heat given out by hot water = heat given out by cold water

QH = Q

mHc (θH - &theta = mcc (θ – θC)

substituting and dividing both sides by c, we have

mHc(90  -  30 ) = mc (30 – 15 )

60mH = 15mc

Saboda an ce mu nemo ratio of cold water to hot water, yanzu zamu maida mc ya zama subject of the equation.

mc = 60mH / 15

yanzu zamu yi rearranging dinsu sai su zama

mc / mH = 60 / 15 = 4 /1

ratio of mass of cold water to mass of hot water is 4:1

idan kuma aka ce muyi lissafin ration of mass of hot water to mass of cold water,you would proceed as follows:

daga 60mH = 15mc, make mH subject of the equation

mH = 15mc / 60

mc / mH = 15 / 60 = 1 / 4

Therefore, ratio of mass of hot water to mass of cold water is 1:4

Wadannan sune kadan daga cikin misalan yadda ake lissafin mixture method na measurement of specific heat capacity. A makala ta mai zuwa zan kawo muku electrical method na specific heat capacity, har ila yau ina kara tuna sar damu akan kokari wajen daukar wasu tambayoyi domin gwada amsansu. Yin hakan zai taimaka wajen kara koyan kowanne lissafi shine zai sa dalibi ya gane cewa lallai ya iya wannan lissafin ko be iya ba. Aci gaba da kasancewa tare damu ta fannin ilmin kimiyyar lissafi (physics) don samun saukin fahimtar kowanne topic, mun gode.

Mai karatu na iya duba:Takaitaccen bayani game da gravitational field

Hakkin mallakar hoto (photo credit): bdyczewski from Pixabay