1. Nkọwa nke ụkpụrụ ndị isi ojii
Ọ bụrụ na anyị agwakọta mmiri ọkụ na oyi n'ime akpa mepere emepe (dịka ọmụmaatụ bọket), okpomọkụ na-esi na mmiri ọkụ gaa na mmiri oyi. Ebe ọ bụ na akpa ahụ mepere emepe, mgbe ahụ okpomọkụ ụfọdụ na-abanye n'ikuku. Akpa ahụ na-adịkwa ọkụ. Okpomọkụ nke mmiri ọkụ na-ewepụta abụghị naanị mmiri oyi na-amịkọrọ, kamakwa ikuku na akpa na-amịkọrọkwa ya. N'ọnọdụ a, bọket ahụ bụ usoro na-abụghị nkewa. Mgbe anyị gwakọtara mmiri ọkụ na oyi na thermos mechiri emechi, okpomọkụ na-esi na mmiri ọkụ gaa na mmiri oyi. Thermos bụ sistemu dịpụrụ adịpụ, yabụ okpomọkụ ọ bụla na-abanye n'ikuku ma ọ bụ thermos. Mmiri oyi na-amịkọrọ naanị okpomọkụ nke mmiri ọkụ na-ewepụta ruo mgbe ngwakọta nke mmiri ọkụ na mmiri oyi ruru nguzozi okpomọkụ.
Ụkpụrụ Ojii na-ekwu na n'ime sistemụ mechiri emechi nke dịpụrụ adịpụ, okpomọkụ nke ihe dị elu na-ewepụta = okpomọkụ nke ihe dị ala na-amịkọrọ.
2. Usoro isi ojii
Mwepụta Q = Ntinye Q
Mwepụta Q = okpomọkụ nke ihe ndị dị elu na-ewepụta, nnabata Q = okpomọkụ nke ihe ndị dị ala na-amịkọrọ.
3. Ihe atụ nke ajụjụ na mkparịta ụka
Ihe atụ nke nsogbu 1.
A na-etinye ígwè ọkụ nke nwere oke nke 1 kg na okpomọkụ nke 100 o C n'ime akpa nwere kilogram mmiri 2 na okpomọkụ nke 20 o C. Kedu okpomọkụ nke ngwakọta ikpeazụ? Okpomọkụ kpọmkwem nke ígwè = 450 J/kg C o , okpomọkụ kpọmkwem nke mmiri = 4200 J/kg C o.
A maara:
Ibu ígwè (m) = 1 kg
Okpomọkụ nke ígwè (T) = 100 oC
Mmiri dị n'ọbara (m) = 2 kg
Okpomọkụ mmiri (T) = 20 oC
Okpomọkụ kpọmkwem maka ígwè (c) = 450 J/kg C o
Okpomọkụ kpọmkwem maka mmiri (c) = 4200 J/kg C o
Achọrọ m: T?
Ngwọta:
Ígwè nwere okpomọkụ dị elu karịa mmiri nke na ígwè na-ewepụta okpomọkụ, mmiri na-amịkọrọ okpomọkụ.
Mwepụta Q = Ntinye Q
mc ΔT = mc ΔT
(1)(450)(100-T) = (2)(4200)(T-20)
(450)(100-T) = (8400)(T-20)
45000 – 450T = 8400 T – 168000
45000 + 168000 = 8400 T + 450 T
213000 = 8850 T
T = 213000: 8850
T = 24 o C
Okpomọkụ ikpeazụ nke ngwakọta nke ígwè ọkụ na mmiri oyi mgbe ha abụọ nọ na nhazi okpomọkụ bụ 24 oC.
Ihe atụ nke nsogbu nke abụọ.
0.2 kg nke ice agwakọtara ya na tii ọkụ nke nwere oke 0.2 kg. Okpomọkụ nke ice = -10 o C, okpomọkụ nke tii ọkụ = 40 o C. Okpomọkụ kpọmkwem nke ice = 2100 J/kg C o , okpomọkụ kpọmkwem nke mmiri = 4200 J/kg C o , okpomọkụ nke ngwakọta maka mmiri = 334000 J/kg. A na-agwakọta ice na tii n'ime sistemu dịpụrụ adịpụ, nke mechiri emechi.
A maara:
Ibu ice (m) = 0.2 kg
Ibu tii (m) = 0.2 kg
Okpomọkụ mmiri kpọmkwem (c) = 4180 J/kg C o
Okpomọkụ kpọmkwem nke ice (c) = 2100 J/kg C o
Okpomọkụ nke njikọta mmiri (L F ) = 334 x 10 3 J/kg
Okpomọkụ nke ice (T ice ) = -10 oC
Okpomọkụ tii (T tii ) = 40 oC
Achọrọ: Ngwakọta T
Ngwọta:
Nzọụkwụ mbụ, mee atụmatụ ọnọdụ ikpeazụ
Okpomọkụ nke mmiri ga-ewepụta iji belata okpomọkụ nke 0.2 kg nke tii ọkụ, site na 40 oC ruo 0 oC
Ntọhapụ Q = mc ∆T
Mwepụta Q = (0.2 kg) (4180 J/Kg C o ) (40 o C – 0 o C)
Mwepụta Q = (0.2 kg) (4180 J/Kg C o ) (40 o C)
Mwepụta Q = 33,440 Joule
Mwepụta Q = 33.44 kJ
A na-eji 0.2 kg nke ice eme ka okpomọkụ dịkwuo elu site na -10 o C ruo 0 o C
Q a na-amịkọrọ = mc ∆T
Q a mịrị amị = (0.2 kg) (2100 J/Kg C o ) (0 o C – (-10 o C))
Q a na-amịkọrọ = (0.2 kg) (2100 J/Kg C o ) (10 o C)
Q a na-amịkọrọ = Joule 4,200
Q a na-amịkọrọ = 4.2 kJ
Okpomọkụ a na-amịkọrọ iji gbazee 0.2 kg nke ice (okpomọkụ dị mkpa iji gbazee ice niile ka ọ ghọọ mmiri)
Njikọ Q = m L F
Njikọ Q = (0.2 kg) (334 x 10 3 J/kg)
Njikọ Q = 66.8 x 10 Joule 3
Njikọ Q = 66.8 kJ
Dabere na ngụkọ ndị dị n'elu, nsonaazụ ya:
Mwepụta Q = 33.44 kJ
Q absorption = 4.2 kJ
Njikọ Q = 66.8 kJ
Mgbe tii na-ewepụta okpomọkụ ruru 33.44 kJ, okpomọkụ tii na-agbanwe site na 40 oC ruo 0 oC . A na-eji ụfọdụ okpomọkụ ewepụtara (ihe dị ka 4.2 kJ) ebuli okpomọkụ ice site na -10 oC ruo 0 oC . 33.44 kJ – 4.2 kJ = 29.24 kJ. Okpomọkụ fọdụrụ = 29.24 kJ. Iji gbazee ice niile ka ọ ghọọ mmiri chọrọ okpomọkụ nke 66.8 kJ. Okpomọkụ fọdụrụ bụ naanị 29.24 kJ.
Na mmechi, a na-eji okpomọkụ nke tii ọkụ na-ewepụta naanị iji bulie okpomọkụ ice site na -10 oC ruo 0 oC ma gbazee ụfọdụ ice. Ụfọdụ ice agbazela ghọọ mmiri; ụfọdụ agbazebeghị. N'oge usoro ịgbanwe ihe siri ike ka ọ bụrụ mmiri mmiri, okpomọkụ anaghị agbanwe. Ya mere, okpomọkụ nke ngwakọta ikpeazụ = 0 oC.
Ihe atụ nke nsogbu nke atọ.
oke tii ọkụ (m) = 0.4 kg,
oke ice (m) = 0.2 kg,
okpomọkụ nke ice = -10 o C
okpomọkụ nke tii ọkụ = 90 o C.
Ọ bụrụ na ha abụọ agwakọta, kedu okpomọkụ nke ngwakọta ikpeazụ?
Okpomọkụ kpọmkwem maka mmiri (c) = 4180 J/kg C o ,
Okpomọkụ kpọmkwem maka ice (c) = 2100 J/kg C o ,
Okpomọkụ nke njikọta mmiri (L F ) = 334 x 103 J/kg
Ngwọta:
Nzọụkwụ mbụ: Mee atụmatụ maka steeti ikpeazụ
Okpomọkụ nke mmiri ga-ewepụta iji belata okpomọkụ nke 0.4 kg nke tii ọkụ, site na 90 o C ruo 0 o C
Ntọhapụ Q = mc ∆T
Mwepụta Q = (0.4 kg) (4180 J/Kg C o ) (90 o C – 0 o C)
Mwepụta Q = (0.4 kg) (4180 J/Kg C o ) (90 o C)
Mwepụta Q = 150,480 Joule
Mwepụta Q = 150.48 kJ
Okpomọkụ nke 0.2 kg nke ice na-amịkọrọ iji mee ka okpomọkụ ya dịkwuo elu site na -10 oC ruo 0 oC
Q a na-amịkọrọ = mc ∆T
Q a mịrị amị = (0.2 kg) (2100 J/Kg C o ) (0 o C – (-10 o C))
Q a na-amịkọrọ = (0.2 kg) (2100 J/Kg C o ) (10 o C)
Q a na-amịkọrọ = Joule 4,200
Q a na-amịkọrọ = 4.2 kJ
Okpomọkụ dị mkpa iji gbazee 0.2 kg nke ice (okpomọkụ dị mkpa iji mee ka ice niile ghọọ mmiri)
Njikọ Q = m L F
Njikọ Q = (0.2 kg) (334 x 10 3 J/Kg)
Njikọ Q = 66.8 x 10 Joule 3
Njikọ Q = 66.8 kJ
Dabere na ngụkọ ndị dị n'elu, nsonaazụ ya:
Mwepụta Q = 150.48 kJ
Q a na-amịkọrọ = 4.2 kJ
Njikọ Q = 66.8 kJ
Mgbe tii ọkụ wepụtara 150.48 kJ nke okpomọkụ, okpomọkụ tii ọkụ ahụ gbanwere site na 90 o C ruo 0 o C. E jiri ụfọdụ okpomọkụ ewepụtara (ihe dị ka 4.2 kJ) bulie okpomọkụ ice site na -10 o C ruo 0 o C.
150.48 kJ – 4.2 kJ = 146.28 kJ. Okpomọkụ fọdụrụ = 146.28 kJ.
Okpomọkụ achọrọ iji gbazee ice niile ka ọ ghọọ mmiri bụ naanị 66.8 kJ. 146.28 kJ – 66.8 kJ = 79.48 kJ. O mechara pụta na ihe fọdụrụ bụ 79.48 kJ. Tii na-ekpo ọkụ anaghị ewepụta okpomọkụ niile ruo mgbe okpomọkụ ahụ belatara ruo 0 oC . Mmechi: okpomọkụ nke ngwakọta ikpeazụ ga-abụrịrị karịa 0 oC.
Nzọụkwụ nke Abụọ: Chọpụta okpomọkụ ikpeazụ (T)
Okpomọkụ achọrọ iji mee ka okpomọkụ ice dịkwuo elu site na -10 o C ruo 0 o C = 4200 Joules
Okpomọkụ ahụ aghaghị ịgbaze ice niile ka ọ bụrụ mmiri nke a makwaara dịka okpomọkụ zoro ezo = Joules 66,800
Okpomọkụ ahụ aghaghị ime ka okpomọkụ nke mmiri (mmiri sitere na ngwakọta ice niile) si na 0 o C ruo T = (oke ice) (okpomọkụ mmiri) (T – 0 o C) = (0.2 kg) (4180 J/kg C o )(T) = (836 T) J/C o
Okpomọkụ nke tii ọkụ na-ewepụta iji belata okpomọkụ site na 90 o C ruo T = (oke mmiri ọkụ) (ụdị okpomọkụ) (90 o C – T) = (0.4 kg) (4180 J/kg C o ) (90 o C – T) = 1672 J/C o (90 o C – T) = 150,480 J – (1672 T) J/C o
4200 J + 66,800 J + (836 T) J/C o = 150,480 J – (1672 T) J/C o
71,000 J + (836 T) J/C o = 150,480 J – (1672 T) J/C o
(836 T) J/C o + (1672 T) J/C o = 150,480 J – 71,000 J
(2508 T) J/C o = 79,480 J
T = 31.7 o C
Okpomọkụ ikpeazụ = 31.7 o C
