Egzanp pwoblèm chalè

5 Egzanp pwoblèm chalè

1. Fèblan ki gen yon mas 2 kilogram genyen chalè espesifik 130 J.kg^-1C^-1Kantite a chalè ke fèblan an absòbe pou ogmante tanperati li soti nan 50^oC rive nan 100^oC se...

Pembahasan

Li konnen ke:

Mas (m) = 2 kg

Chalè espesifik eten (c) = 130 J.kg^-1C^-1

Chanjman tanperati (ΔT) = 100^oC - 50^oC = 50^oC

Mande: Chalè (Q) absòbe pa eten

Repons:

Fòmil chalè:

Q = mc ΔT

Deskripsyon: Q = chalè, m = mas, c = chalè espesifik, ΔT = chanjman suhu

Chalè absòbe pa fèblan:

Q = (2 kg)(130 J.kg^-1C^-1) (50^oC)

Q = (100)(130)

Q = 13000 Joule

Q = 1,3 x 10⁴ Joule

2. Kuiv ak yon chalè espesifik 390 J kg^{-1 o}C^-1 te fè eksperyans yon chanjman tanperati 40^oC. Si chalè kwiv la absòbe pandan chanjman tanperati li a se 200 kalori, alò mas kwiv la se...

Pembahasan

Li konnen ke:

Chalè espesifik kwiv (c) = 390 J kg^{-1 o}C^-1

Chanjman tanperati (ΔT) = 40^oC

Chalè (Q) = 40 Joules

Mande: Mas (m) kwiv

Repons:

Q = mc ΔT

40 = (m)(390) (40^oC)

40 = (mèt)(390)(40)

40 = (m)(390)(4)

40 = (m)(1560)

m = 40 / 1560

m = 0,026 kg

Mas kwiv la se 26 gram.

3. Dlo gen yon mas 20 gram ak yon chalè espesifik 1 kal g^{-1 o}C^-1 okòmansman nan yon tanperati 30^oC. Detèmine chanjman nan tanperati dlo a si kantite chalè dlo a absòbe a se 300 kalori!

Pembahasan

Li konnen ke:

Mas (m) = 20 gr

Tanperati inisyal (T₁) = 30^oC

Chalè espesifik dlo (c) = 1 kal g^{-1 o}C^-1

Chalè (Q) = 300 kalori

Mande: Chanjman nan tanperati dlo a

Repons:

Q = mc ΔT

Q = mc (T₂ - T₁)

300 kalori = (20 gram)(1 kalori g^{-1 o}C^-1)(T₂ - 30^oC)

300 = (20)(1)(T₂ - 30)

300 = 20 (T₂ - 30)

300 = 20 T₂ - 600

300 + 600 = 20 T₂

900 = 20 T₂

T₂ = 900/20

T₂ = 45^oC

Chanjman tanperati:

ΔT = 45^oC - 30^oC

ΔT = 15^oC

4. Kantite chalè ki nesesè pou bese tanperati 4 kg likid ki gen yon chalè espesifik 500 J/kg K soti nan 40 ^oC vin tounen 20 ^oC se…
A. 20 kJ
B. 30 kJ
C. 40 kJ
D. 50 kJ
E. 60 kJ
Diskisyon:
Li konnen :
Mas (m) = 4 kg
Chalè espesifik (c) = 500 J/kg K = 500 J/kg C^o
Chanjman tanperati (delta T) = 40 ^oC - 20 ^oC = 20 C^o
Mande :
Chalè (Q)
Jawab :
Fòmil Chalè :
Q = (m)(c)(delta T)
Q = (4)(500)(20) = 40.000 Joul = 40 kJ
kJ = kilo Joule
Repons ki kòrèk la se C.

5. Yon objè solid ki gen yon mas 2 kg chofe apati yon tanperati 30^oC vin tounen 100^oC, mande 7 × 10 chalè⁵ Joule. Chalè espesifik objè a se…
A. 400 J/kg K
B. 500 J/kg K
C. 700 J/kg K
D. 1000 J/kg K
E. 5000 J/kg K
Diskisyon:
Li konnen :
Mas (m) = 2 kg
Chalè (Q) = 7 × 10⁵ J = 700.000 J
Chanjman tanperati (delta T) = 100 ^oC - 30 ^oC = 70 C^o
Mande :
Chalè espesifik (c) ?
Jawab :
c = Q / m T
c = 700.000 J / (2 kg)(70 C^o) = 700.000 J / 140 kg C^o
c = 5000 J/kg C^o
c = 5000 J/kg K
Repons ki kòrèk la se E.