Okpomọkụ 9 nke a kapịrị ọnụ nke oke okpomọkụ Mgbanwe na okpomọkụ - Nsogbu na Ngwọta
1. A na-ekpo ọkụ otu kilogram 2 site na 50oC ka 100oC. Ihe kpọmkwem ọkụ nke ndu bụ 130 J.kg-1 oC-1. Ego ole okpomọkụ ndu ahụ ọ̀ na-amịkọrọ ya?
A maara:
Ibu (m) = 2 kg
Okpomọkụ kpọmkwem (c) = 1 3 0 J.kg -1 C -1
Mgbanwe na okpomọkụ ( Δ T) = 100 o C – 50 o C = 50 o C
Achọtara: Okpomọkụ (Q)
Ngwọta:
Q = mc Δ T
Q = okpomọkụ , m = oke, c = okpomọkụ kpọmkwem , Δ T = mgbanwe na okpomọkụ
Okpomọkụ nke lead na-amịkọrọ:
Q = (2 kg)(1 3 0 J.kg -1 C -1 )(50 o C)
Q = (100)(1 3 0)
Q = 1 3,000 Joules
Q = 1.3 x 104 Joules
[irp]
2. Okpomọkụ kpọmkwem nke ọla kọpa bụ 390 J/k g o C, mgbanwe na okpomọkụ bụ 40 o C. Ọ bụrụ na ọla kọpa ahụ na-amịkọrọ okpomọkụ 40 Joules, gịnị bụ oke ọla kọpa ahụ!
A maara:
Okpomọkụ kpọmkwem nke ọla kọpa (c) = 390 J/kg g o C
Mgbanwe na okpomọkụ ( ΔT ) = 40 oC
Okpomọkụ ( Q) = 40 J
Achọtara: Oke (m) nke ọla kọpa
Ngwọta:
Q = mc Δ T
40 J = (m)( 390 J/k g o C)(40 o C)
40 = (m)( 390 /kg g)(40)
40 = (m)( 390 /kg )(4)
40 = (m)( 1560 /kg )
m = 40 / 1560
m = 0.026 kg
m = gram 26
3. Okpomọkụ mbụ nke gram 20 nke mmiri bụ 30 oC . Okpomọkụ kpọmkwem nke mmiri bụ 1 cal g -1 oC -1 . Ọ bụrụ na mmiri amịkọrọ kalori 300 nke okpomọkụ, chọpụta okpomọkụ ikpeazụ!
A maara:
Ibu (m) = 20 g
Okpomọkụ mbụ ( T 1 ) = 30 oC
Okpomọkụ kpọmkwem nke mmiri (c) = 1 kalori gr -1 o C -1
Okpomọkụ (Q) = 300 kalori
Achọrọ: Okpomọkụ ikpeazụ nke mmiri
Ngwọta:
Q = mc ΔT
300 kalori = (20 gr)( 1 kalori gr -1 o C -1 )( T 2 -30 )
300 = (20)(1)( T 2 -30)
300 = 20 ( T 2 -30 )
300 = 20 T 2 – 600
300 + 600 = 20T 2
900 = 20T 2
T 2 = 900 / 20
T 2 = 45
Mgbanwe nke okpomọkụ bụ 45 o C – 30 o C = 15 o C.
[irp]
4. Tmgbanwe na okpomọkụ nke mmiri oké osimiri is 1oC mgbe mmiri na-amịkọrọ okpomọkụ 3900 joules. Okpomọkụ kpọmkwem nke mmiri oké osimiri bụ 3.9 × 103 J/kg°C, kedu oke mmiri dị n'oké osimiri?
A maara:
Mgbanwe na okpomọkụ ( ΔT ) = 1 oC
Okpomọkụ (Q) = Joule 3900
Okpomọkụ kpọmkwem nke mmiri oké osimiri (c) = 3.9 x 10 3 J/ k g°C = 3900 J/Kg°C
Achọtara: Oke (m)
Ngwọta:
Q = mc Δ T
Q = okpomọkụ , m = oke, c = okpomọkụ kpọmkwem , Δ T = mgbanwe na okpomọkụ
m = Q / c Δ T = 3900 / (3900)(1) = 3900 / 3900 = 1 kg
5. Ọla kọpa dị kilogram 2 na-amịkọrọ okpomọkụ 39,000 J na 30°C. Ọ bụrụ na okpomọkụ ọla kọpa kpọmkwem bụ 390 J/kg °C, kedu okpomọkụ ikpeazụ nke ọla kọpa ahụ...
A maara:
Ibu (m) = 2 kg
Okpomọkụ mbụ ( T 1 ) = 30 oC
Okpomọkụ (Q) = Joule 39,000
Okpomọkụ kpọmkwem (c) nke ọla kọpa = 390 J/kg o C
Achọrọ : Okpomọkụ ikpeazụ ( T2 )
Ngwọta:
Q = mc ΔT
Q = okpomọkụ , m = oke, c = okpomọkụ kpọmkwem , ΔT = mgbanwe na okpomọkụ
Q = mc ΔT = mc (T 2 – T 1 )
39,000 = (2)( 390)(T2 – 30)
100 = (2)(1)(T 2 – 30)
100 = (2)(T 2 – 30)
50 = T 2 – 30
T 2 = 50 + 30
T 2 = 80 o C
[irp]
6. A na-ekpo ọkụ mmiri dị kilogram 5 site na 15°C ruo 40°C. Gịnị bụ okpomọkụ mmiri na-amịkọrọ. Okpomọkụ kpọmkwem nke mmiri bụ 4.2 × 10 3 J/Kg° C.
A maara:
Ibu (m) = 5 kg
Okpomọkụ mbụ (T 1 ) = 15°C
Okpomọkụ ikpeazụ (T2 ) = 40°C
Okpomọkụ mmiri kpọmkwem (c) = 4.2 × 10 3 J/kg° C
Achọtara: Okpomọkụ (Q)
Ngwọta:
Q = mc ΔT
Q = (5 kg)( 4.2 × 10 3 J/kg°C)(40°C – 15°C)
Q = (5)( 4.2 × 10 3 J)(25)
Q = 525 x 10 3 J
Q = 525,000 Joule
7. A na-ekpo ọkụ mmiri dị kilogram 2 site na 24°C ruo 90°C. Gịnị bụ okpomọkụ mmiri na-amịkọrọ. Okpomọkụ kpọmkwem nke mmiri bụ 4.2 × 10 3 J/Kg° C.
A maara:
Ibu (m) = 2 kg
Okpomọkụ mbụ (T 1 ) = 24°C
Okpomọkụ ikpeazụ (T2 ) = 90°C
Okpomọkụ mmiri kpọmkwem (c) = 4,200 Joules/kg°C
Achọtara m :: Okpomọkụ (Q)
Ngwọta:
Q = m c ΔT
Q = (2 kg)( 4,200 Joules/kg°C)(90°C – 24°C)
Q = (2 kg)( 4,200 Joules/kg°C)(66°C)
Q = (132)( 4,200 Joules)
Q = 554,400 Joule
[irp]
8. A na-ekpo ọkụ mmiri gram 5 site na 10°C ruo 40°C. Gịnị bụ okpomọkụ mmiri na-amịkọrọ. Okpomọkụ kpọmkwem nke mmiri bụ 1 × 10 3 cal /g r ° C.
A maara:
Ibu (m) = gram 5
Okpomọkụ mbụ ( T 1 ) = 10 oC
Okpomọkụ ikpeazụ (T2 ) = 40 oC
Okpomọkụ mmiri kpọmkwem (c) = 1 c al/ gr°C
Achọtara : Okpomọkụ
Ngwọta:
Q = mc ΔT
Q = (gram 5)( 1 kalori/gr°C)(40 o C – 10 o C)
Q = (5)( 1 c al)(30)
Q = kalori 150
9. Mmiri dị kilogram 0.2 na-amịkọrọ okpomọkụ ruru 42,000 na 25 oC . Okpomọkụ mmiri kpọmkwem bụ 4200 J/kg oC , nke bụ okpomọkụ ikpeazụ nke mmiri.
A maara:
Mmiri dị n'ọbara (m) = 0.2 kg
Okpomọkụ (Q) = Joule 42,000
Okpomọkụ mmiri kpọmkwem (c) = 4200 J/kg o C
Okpomọkụ mbụ ( T 1 ) = 25 oC
Achọrọ: Okpomọkụ ikpeazụ ( T2 )
Ngwọta:
Q = mc ΔT = mc (T 2 – T 1 )
Ajụjụ = okpomọkụ, m = oke, c = kpọmkwem ọkụ, ΔT = mgbanwe na okpomọkụ, T1 = okpomọkụ mbụ, T2 = okpomọkụ ikpeazụ
Q = mc (T 2 – T 1 )
42,000 = (0.2)(4200)(T 2 – 25)
42,000 = 840 (T 2 – 25)
42,000 = 840 T 2 – 21,000
42,000 + 21,000 = 840 T 2
63,000 = 840 T 2
T 2 = 63,000 / 840
T 2 = 75 o C
- Ịgbanwe nha okpomọkụ
- Mgbasawanye ahịrị
- Mgbasawanye mpaghara
- Mgbasawanye olu
- okpomọkụ
- Ihe nhata igwe nke okpomọkụ
- Ike okpomọkụ na okpomọkụ kpọmkwem
- Okpomọkụ zoro ezo, okpomọkụ nke njikọta, okpomọkụ nke vaporization
- Nchekwa ike maka nnyefe okpomọkụ
