Panas Massa Panas Spesifik Owah-owahan suhu – Masalah lan Solusi

9 Panas Massa Panas spesifik Owah-owahan suhu – Masalah lan Solusi

1. Timah 2 kg dipanasake saka 50^oC kanggo 100^oC. Ing panas spesifik timbal yaiku 130 J.kg^{-1 o}C^-1. Pinten panas diserap dening timah?

Dikenal:

Massa (m) = 2 kg

Panas spesifik (c) = 130 J.kg^-1C^-1

Owah-owahan suhu (ΔT) = 100^oC - 50^oC = 50^oC

Dikarepake: panas (Q)

Solusi:

Q = mc ΔT

Q = panas, m = massa, c = panas tartamtu, ΔT= owah-owahan suhu

Panas sing diserep dening timbal:

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

Q = (100)(1)30)

Q = 13,000 jro

Q = 1.3 x 10⁴ Joule

2. Panas spesifik tembaga yaiku 390 J/kg ^oC, owah-owahan suhu yaiku 40^oC. Yen tembaga nyerep panas 40 Joule, pira massa tembaga kasebut!

Dikenal:

Panas spesifik tembaga (c) = 390 J/kg^oC

Owah-owahan suhu (ΔT) = 40^oC

Panas (Q) = 40 J

Dikarepake: Massa (m) saka tembaga

Solusi:

Q = mc ΔT

40 J = (m)(390 J/kg ^oC)(40^oC)

40 = (m)(390/kg)(40)

40 = (m)(390/kg)(4)

40 = (m)(1560/kg)

m = 40 / 1560

m = 0.026 kg

m = 26 gram

3. Suhu awal 20 gram banyu yaiku 30^oC. Kalor spesifik banyu yaiku 1 kal g^{-1 o}C^-1Yen banyu nyerep 300 kalori panas, temtokake suhu pungkasan!

Dikenal:

Massa (m) = 20 gr

Suhu awal (T₁) = 30^oC

Panas spesifik banyu (c) = 1 kal gr^{-1 o}C^-1

panas (Q) = 300 kal

Dikarepake: Suhu pungkasan banyu

Solusi:

Q = mc T

300 kal = (20 gr)(1 kal g^{-1 o}C^-1)(T₂-30)

300 = (20)(1)(T₂-30)

300 = 20 (T₂-30)

= 300 20T₂ - 600

300 + 600 = 20T₂

900 = 20T₂

T₂ = 900/20

T₂ = 45

Owah-owahan suhu yaiku 45^oC - 30^oC = 15^oC.

4. Towah-owahan suhu saka banyu segara is 1^oC nalika banyu nyerep 3900 Joule panas. Panas spesifik banyu segara yaiku 3.9 × 10³ J/kg°C, pira bobote banyu segara kuwi.

Dikenal:

Owah-owahan suhu (ΔT) = 1^oC

panas (Q) = 3900 Joule

Panas spesifik banyu segara (c) = 3.9 x 10³ J/kg°C = 3900 J/Kg°C

Dikarepake: Massa (M)

Solusi:

Q = mc ΔT

Q = panas, m = massa, c = panas spesifik, ΔT= owah-owahan suhu

m = Q / c ΔT = 3900 / (3900)(1) = 3900 / 3900 = 1 kg

5. Tembaga 2 kg nyerep panas 39,000 J ing suhu 30°C. Yen panas spesifik tembaga yaiku 390 J/kg °C, pira suhu pungkasan tembaga kasebut…

Dikenal:

Massa (m) = 2 kg

Suhu awal (T₁) = 30^oC

panas (Q) = 39,000 Joule

Panas tartamtu (c) saka tembaga = 390 J/kg ^oC

wanted : Suhu pungkasan (T₂)

Solusi:

Q = mc T

Q = panas, m = massa, c = panas spesifik, ΔT = owah-owahan suhu

Q = mc ΔT = mc (T₂ - T₁)

39,000 = (2)(390)(T₂ - 30)

100 = (2)(1)(T₂ - 30)

100 = (2)(T₂ - 30)

50 = T₂ - 30

T₂ = 50 + 30

T₂ = 80^oC

6. Banyu 5 kg dipanasake saka suhu 15°C nganti 40°C. Pira panas sing diserap banyu. Panas spesifik banyu yaiku 4.2 × 10³ J/Kg°C.

Dikenal:

Massa (m) = 5 kg

Suhu awal (T₁) = 15°C

Suhu pungkasan (T₂) = 40°C

Panas spesifik banyu (c) = 4.2 × 10³ J/kg°C

Dikarepake: panas (Q)

Solusi:

Q = mc T

Q = (5 kg)(4.2 × 10³ J/kg°C)(40°C – 15°C)

Q = (5)(4.2 × 10³ J)(25)

Q = 525 x 10³ J

Q = 525,000 Joule

7. Banyu 2 kg dipanasake saka suhu 24°C nganti 90°C. Pira panas sing diserap banyu. Panas spesifik banyu yaiku 4.2 × 10³ J/Kg°C.

Dikenal:

Massa (m) = 2 kg

Suhu awal (T₁) = 24°C

Suhu pungkasan (T₂) = 90°C

Panas spesifik banyu (c) = 4,200 Joule/kg°C

Dikarepake :: panas (Q)

Solusi:

Q = m c T

Q = (2 kg)(4,200 Joule/kg°C)(90°C – 24°C)

Q = (2 kg)(4,200 Joule/kg°C)(66°C)

Q = (132)(4,200 Joule)

Q = 554,400 Joule

8. Banyu 5 gram dipanasake saka suhu 10°C nganti 40°C. Pira panas sing diserap banyu. Panas spesifik banyu yaiku 1 × 10³ cal/gr°C.

Dikenal:

Massa (m) = 5 gram

Suhu awal (T₁) = 10^oC

Suhu pungkasan (T₂) = 40^oC

Panas spesifik banyu (c) = 1 cal/ gr°C

wanted : Panas

Solusi:

Q = mc T

Q = (5 gram)(1 kal/gr°C)(40^oC - 10^oC)

Q = (5)(1 cal)(30)

Q = 150 Kalori

9. Banyu 0.2 kg nyerep 42,000 Joule panas ing 25^oC. Panas spesifik banyu yaiku 4200 J/kg ^oC, pira suhu pungkasan banyu.

Dikenal:

Massa banyu (m) = 0.2 kg

panas (Q) = 42,000 Joule

Panas spesifik banyu (c) = 4200 J/kg ^oC

Suhu awal (T₁) = 25^oC

Dikarepake: Suhu pungkasan (T₂)

Solusi:

Q = mc ΔT = mc(T)₂ - T₁)

Q = panas, m = massa, c = panas spesifik, ΔT = owah-owahan suhu, T₁ = suhu awal, T₂ = suhu pungkasan

Q = mc(T₂ - T₁)

42,000 = (0.2)(4200)(T₂ - 25)

42,000 = 840 (T₂ - 25)

42,000 = 840 T₂ - 21,000

42,000 + 21,000 = 840 T₂

63,000 = 840 T₂

T₂ = 63,000/840

T₂ = 75^oC

1. Ngonversi skala suhu
2. Ekspansi linier
3. Ekspansi wilayah
4. Ekspansi volume
5. panas
6. Padanan mekanik saka panas
7. Panas spesifik lan kapasitas panas
8. Panas laten, panas fusi, panas penguapan
9. Konservasi energi kanggo transfer panas