Basic Physics

# Black principle – problems and solutions

1. 200-gram water at 30°C mixed with 100-gram water at 90°C. The specific heat of water = 1 cal.gram−1°C−1. Determine the final temperature of the mixture!

Known :

Mass of water 2 (m1) = 200 gram

Temperature of water 1 (T1) = 30oC

Mass of water 2 (m2) = 100 gram

Temperature of water 2 (T2) = 90oC

The specific heat for water (c) = 1 cal.gram−1°C−1

Wanted : The final temperature

Solution :

Heat released by hooter water (Q2) = heat absorbed by cooler water (Q1)

m2 c (ΔT) = m1 c (ΔT)

(100)(1)(90 – T) = (200)(1)(T – 30)

(100)(90 – T) = (200)(T – 30)

9000 – 100T = 200T – 6000

9000 + 6000 = 200T + 100T

15000 = 300T

T = 15000/300

T = 50oC

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2. 60-gram at 90oC mixed with 40-gram water at 25oC. What is the final temperature of the mixture. The specific heat for water = 1 cal.g-1.oC-1.

Known :

Mass of water 1 (m1) = 60 gram

Temperature of water 1 (T1) = 90oC

Mass of water 2 (m2) = 40 gram

Temperature of water 2 (T2) = 25oC

The specific heat for water (c) = 1 cal.g-1.oC-1

Wanted : The final temperature

Solution :

Heat released by hooter water (Q2) = heat absorbed by cooler water (Q1)

m1 c (ΔT) = m2 c (ΔT)

(60)(1)(90 – T) = (40)(1)(T – 25)

(60)(90 – T) = (40)(T – 25)

5400 – 60T = 40T – 1000

5400 + 1000 = 40T + 60T

6400 = 100T

T = 6400/100

T = 64oC

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3. M-gram ice at 0oC placed in 340-gram water at 20oC in a specific container. Latent heat of fusion (Lice) = 80 cal g-1, the specific heat of water (cwater) = 1 cal g-1 oC-1. All the ice melts and the thermal equilibrium = 5oC. Find mass of ice.

Known :

Mass of water (m) = 340 gram

Temperature of ice (Tice) = 0oC

Temperature of water (Twater) = 20oC

Temperature of thermal equilibrium (T) = 5oC

Latent heat of fusion for ice (Lice) = 80 cal g-1

The specific heat for water (cwater) = 1 cal g-1 oC-1

Wanted : Mass of ice (M)

Solution :

Heat released by water (Q2) = heat absorbed by ice (Q1)

mwater cwater (ΔT) = mice Lice + mice cwater (ΔT)

(340)(1)(20-5) = M (80) + M (1)(5-0)

(340)(15) = 80M + 5M

5100 = 85M

M = 5100/85

M = 60 gram

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4. A copper at 100oC placed in 128-gram of water at 30 oC. The specific heat for water is 1 cal.g-1oC-1 and the specific heat for copper is 0.1 cal.g-1oC-1. If the temperature of the thermal equilibrium = 36 oC, what is the mass of copper.

Known :

Temperature of copper (T1) = 100 oC

The specific heat for copper (c1) = 0.1 cal.g-1oC-1

Mass of water (m2) = 128 gram

Temperature of water (T2) = 30 oC

The specific heat for water (c2) = 1 cal.g-1oC-1

The temperature of the thermal equilibrium (T) = 36 oC

Wanted : Mass of copper (m1)

Solution :

Q copper = Q water

m1 c1 ΔT = m2 c2 ΔT

(m1)(0.1)(100-36) = (128)(1)(36-30)

(m1)(0.1)(64) = (128)(1)(6)

(m1)(6.4) = 768

m1 = 768 / 6.4

m1 = 120 gram

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5. A 3-kg lead with the specific heat for lead is 1400 J.kg-1C-1 at 80oC placed in 10-kg water with the specific heat for water is 4200 J.kg-1C-1. The temperature of the thermal equilibrium is 20oC. What is the initial temperature of water.

Known :

Mass of lead (m1) = 3 kg

The specific heat for lead (c1) = 1400 J.kg-1C-1

Temperature of lead (T1) = 80 oC

Mass of water (m2) = 10 kg

The specific heat of water (c2) = 4200 J.kg-1C-1

The temperature of the thermal equilibrium (T) = 20 oC

Wanted : The initial temperature of water (T2)

Solution :

Q release = Q absorb

m1 c1 ΔT = m2 c2 ΔT

(3)(1400)(80-20) = (10)(4200)(20-T)

(4200)(60) = (42,000)(20-T)

252,000 = 840,000 – 42,000 T

42,000 T = 840,000 – 252,000

42,000 T = 588,000

T = 588,000 / 42,000

T = 14

The initial temperature of water is 14oC.

6. 75-gram water at 0oC mixed with 50-gram water so the temperature of mixture is 40oC. What is the initial temperature of 50-gram water.

Known :

Mass of water 1 (m1) = 75 gram

The initial temperature of water 1 (T1) = 0oC

Mass of water 2 (m2) = 50 gram

The temperature of mixture (T) = 40oC

Wanted : The initial temperature of water 2 (T2)

Solution :

Heat released by hooter water (Qrelease) = heat absorbed by cooler water (Qabsorbs)

m1 c (ΔT1) = m2 c (ΔT2)

m1 (ΔT1) = m2 (ΔT2)

(75)(40 – 0) = (50)(T2 – 40)

(75)(40) = (50)(T2 – 40)

3000 = 50 T2 2000

3000 + 2000 = 50 T2

5000 = 50 T2

T2 = 100 oC

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7. 200-gram metal heated to 120oC, then placed in 100-gram water at 30oC. The temperature of mixture at thermal equilibrium is 60oC. If the specific heat of water is 4200 J.kg-1 oC-1, what is the specific heat of the metal.

Solution :

Convert mass unit from gram to kilogram (International unit)

Known :

Mass of metal (m1) = 200 gram = 0.2 kg

Temperature of metal (T1) = 120oC

Mass of water (m2) = 100 gram = 0.1 kg

Temperature of water (T2) = 30oC

The temperature of mixture (T) = 60oC

The specific heat of water (c2) = 4200 J.kg-1 oC-1

Wanted : The specific heat of metal (c1)

Solution :

Qrelease = Qabsorb

m1 c1 (ΔT1) = m2 c2 (ΔT2)

(0.2)(c1)(120 – 60) = (0.1)(4200)(60 – 30)

(0.2)(c1)(60) = (0.1)(4200)(30)

12 c1 = 12600

c1 = 12600 / 12

c1 = 1050

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