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

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

1. A 2 kg lead is heated from 50^oC līdz 100^oC. īpašs karstums of lead is 130 J.kg^{-1 o}C^-1. Cik daudz siltums is absorbed by the lead?

Zināms:

Masa (m) = 2 kg

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

Temperatūras izmaiņas (ΔT) = 100^oC - 50^oC = 50^oC

Meklē: Siltums (Q)

šķīdums:

Q = mc ΔT

Q= siltums, m = mass, c = the specific heat, ΔT = temperatūras izmaiņas

The heat absorbed by lead :

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

Q = (100)(130)

Q = 13,000 džouls

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!

Zināms:

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

Temperatūras izmaiņas (ΔT) = 40^oC

Heat (Q) = 40 J

Meklē: Masa (m) no vara

šķīdums:

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!

Zināms:

Mass (m) = 20 gr

Sākotnējā temperatūra (T₁) = 30^oC

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

Siltums (Q) = 300 cal

Meklē: The final temperature of water

šķīdums:

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.

Zināms:

Temperatūras izmaiņas (ΔT) = 1^oC

Siltums (Q) = 3900 Joule

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

Meklē: Masa (Kungs)

šķīdums:

Q = mc ΔT

Q= siltums, m = mass, c = īpašs karstums, ΔT = temperatūras izmaiņas

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…

Zināms:

Masa (m) = 2 kg

Sākotnējā temperatūra (T₁) = 30^oC

Siltums (Q) = 39,000 Joule

Īpašs karstums (C) no vara = 390 J/kg ^oC

Wanted : Galīgā temperatūra (T₂)

šķīdums:

Q = mc ΔT

Q= siltums, m = mass, c = īpašs karstums, ΔT = temperatūras izmaiņas

Q = mc₀ ΔT = mc₀ (T₂ - T.₁)

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.

Zināms:

Masa (m) = 5 kg

Sākotnējā temperatūra (T₁) = 15°C

Final temperature (T₂) = 40°C

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

Meklē: Siltums (Q)

šķīdums:

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 džoulu

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.

Zināms:

Masa (m) = 2 kg

Sākotnējā temperatūra (T₁) = 24°C

Final temperature (T₂) = 90°C

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

Wanted :: Siltums (Q)

šķīdums:

Q = m c ΔT

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

Q = (2 kg)(4,200 džouli/kg°C) (66°C)

Q = (132)(4,200 džouli)

Q = 525 000 džoulu

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 kaļķi/gr° C.

Zināms:

Mass (m) = 5 gram

Sākotnējā temperatūra (T₁) = 10^oC

Final temperature (T₂) = 40^oC

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

Wanted : Siltums

šķīdums:

Q = mc ΔT

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

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

Q = 150 kalorijas

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.

Zināms:

Masa ūdens (m) = 0.2 kg

Siltums (Q) = 42,000 Joule

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

Sākotnējā temperatūra (T₁) = 25^oC

Meklē: Final temperature (T₂)

šķīdums:

Q = mc ΔT = mc(T₂ - T.₁)

Q= siltums, m = mass, c = īpašs karstums, ΔT = temperatūras izmaiņas, T₁ = the initial temperature, T₂ = the final temperature

Q = mc(T₂ - T.₁)

42,000 = (0.2)(4200)(T₂ - 25)

42 000 = 840 (T₂ - 25)

42 000 = 840 tonnas₂ - 21,000

42 000 + 21 000 = 840 tonnu₂

42 000 = 840 tonnas₂

T₂ = 63,000/840

T₂ = 75^oC

1. Temperatūras skalu konvertēšana
2. Lineāra izplešanās
3. Teritorijas paplašināšana
4. Skaļuma paplašināšana
5. Siltums
6. Siltuma mehāniskais ekvivalents
7. Īpatnējais siltums un siltumietilpība
8. Latentais siltums, kušanas siltums, iztvaikošanas siltums
9. Enerģijas saglabāšana siltuma pārnesē