Haba Jisim Haba tentu Perubahan suhu – Masalah dan Penyelesaian

9 Heat Mass Specific heat The change in temperature – Problems and Solutions

1. A 2 kg lead is heated from 50^oC ke 100^oC. The haba tentu of lead is 130 J.kg^{-1 darjah}C^-1. Berapa banyak haba is absorbed by the lead?

Diketahui:

Massa (m) = 2 kg

The specific heat (c) = 130 J.kg^-1C^-1

Perubahan suhu (ΔT) = 100^oC - 50^oC = 50^oC

Dikehendaki: Haba (Q)

Penyelesaian:

Q = mc ΔT

Q= haba, m = mass, c = the specific heat, ΔT = perubahan suhu

The heat absorbed by lead :

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

Q = (100)(130)

S = 13,000 joule

Q = 1.3 x 10⁴ Joule

2. The specific heat of copper is 390 J/kg ^oC, the change in temperature is 40^oC. If the copper absorbs 40 Joule of heat, what is the copper’s mass!

Diketahui:

The specific heat of copper (c) = 390 J/kg^oC

Perubahan suhu (ΔT) = 40^oC

Heat (Q) = 40 J

Dikehendaki: Jisim (m) tembaga

Penyelesaian:

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. The initial temperature of 20 gram water is 30^oC. The specific heat of water is 1 cal g^{-1 o}C^-1. If water absorbs 300 calories of heat, determine the final temperature!

Diketahui:

Mass (m) = 20 gr

Suhu awal (T₁) = 30^oC

The specific heat of water (c) = 1 cal gr^{-1 o}C^-1

Haba (Q) = 300 cal

Dikehendaki: The final temperature of water

Penyelesaian:

Q = mc ΔT

300 cal = (20 gr)(1 cal gr^{-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

The change in temperature is 45^oC - 30^oC = 15^oC.

4. The change in temperature of the sea water is 1^oC when water absorbs 3900 Joule of heat. The specific heat of the sea water is × 3.9 10³ J/kg°C, what is the mass of the sea water.

Diketahui:

Perubahan suhu (ΔT) = 1^oC

Haba (Q) = 3900 Joule

The specific heat of the sea water (c) = 3.9 x 10³ J/kg°C = 3900 J/Kg°C

Dikehendaki: Massa (M)

Penyelesaian:

Q = mc ΔT

Q= haba, m = mass, c = haba tentu, ΔT = perubahan suhu

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

5. A 2-kg copper absorbs 39,000 J of heat at 30°C. If the specific heat of copper is 390 J/kg °C, what is the final temperature of the copper…

Diketahui:

Jisim (m) = 2 kg

Suhu awal (T₁) = 30^oC

Haba (Q) = 39,000 Joule

Haba tentu (C) tembaga = 390 J/kg ^oC

Dikehendaki : Suhu akhir (T₂)

Penyelesaian:

Q = mc ΔT

Q= haba, m = mass, c = haba tentu, ΔT = perubahan 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. A 5-kg water is heated from 15°C to 40°C. What is the heat is absorbed by water. Specific heat of water is × 4.2 10³ J/Kg° C.

Diketahui:

Jisim (m) = 5 kg

Suhu awal (T₁) = 15°C

Suhu akhir (T₂) = 40°C

Specific heat of water (c) = 4.2 × 10³ J/kg°C

Dikehendaki: Haba (Q)

Penyelesaian:

Q = mc ΔT

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. A 2-kg water is heated from 24°C to 90°C. What is the heat is absorbed by water. Specific heat of water is × 4.2 10³ J/Kg° C.

Diketahui:

Jisim (m) = 2 kg

Suhu awal (T₁) = 24°C

Suhu akhir (T₂) = 90°C

Specific heat of water (c) = 4,200 Joule/kg°C

Wanted :: Haba (Q)

Penyelesaian:

Q = m c ΔT

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. A 5-gram water is heated from 10°C to 40°C. What is the heat is absorbed by water. Specific heat of water is 1 × 10³ kapur/gr° C.

Diketahui:

Jisim (m) = 5 gram

Suhu awal (T₁) = 10^oC

Suhu akhir (T₂) = 40^oC

Specific heat of water (c) = 1 cal/ gr°C

Dikehendaki : Haba

Penyelesaian:

Q = mc ΔT

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

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

S = 150 kalori

9. A 0.2-kg water absorbs 42,000 Joule of heat at 25^oC. The specific heat of water is 4200 J/kg ^oC, what is the final temperature of water.

Diketahui:

Massa air (m) = 0.2 kg

Haba (Q) = 42,000 Joule

Specific heat of water (c) = 4200 J/kg ^oC

Suhu awal (T₁) = 25^oC

Dikehendaki: Suhu akhir (T₂)

Penyelesaian:

Q = mc ΔT = mc(T₂ - T₁)

Q= haba, m = mass, c = haba tentu, ΔT = perubahan suhu, T₁ = the initial temperature, T₂ = the final temperature

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. Menukar skala suhu
2. Pengembangan linear
3. Perluasan kawasan
4. Pengembangan volum
5. Haba
6. Setara mekanikal haba
7. Haba tentu dan muatan haba
8. Haba pendam, haba pelakuran, haba pengewapan
9. Penjimatan tenaga untuk pemindahan haba