Exercise Temperature and Heat

Exercise Science

1. When given 30 kJ of heat energy, the temperature of a 4.0 kg block of metal metal increases from 60 ⁰C to 80 ⁰C. What is the specific heat capacity of the metal?

Answer

Know: Q = 30 KJ = 30.000 KJ m = 4,0 Kg

𝑇₁= 60℃

𝑇₂ = 80℃

𝜃 = 𝑇₂− 𝑇₁ = 80℃ − 60℃ = 20℃

Unknow: c = ⋯ ? c = ^𝑄

𝑚.𝜃 = ^30.000𝐽

4,0 𝐾𝑔×20℃= 375𝐽

⁄𝐾𝑔℃

2. A fahrenheit thermometer scale will show three times the value of a celsius thermometer scale at temperatures….

Answer

Know: ℉ = 3℃

℉ = 𝑥

℃ = 𝑦 𝑥 = 3𝑦

℉−IP

SP−IP= ^℃−IP

SP−IP

x−32°

212°−32°= ^y−0°

100°−0°

3y−32°

180° = ^y

3y−32° 100°

9 =^y

5

15𝑦 − 160° = 9𝑦 15𝑦 − 9𝑦 = 160°

𝑦 =^160°

6 = 26,7°

3. 1 kg of water with a temperature of 80 ⁰C is poured into a vessel of aluminum with a mass of 2 kg.

If the initial temperature of the vessel is 30 ⁰C, the specific heat of aluminum is 900 J/kg⁰C, and the specific heat of water is 4,200 J/kg⁰C, then determine the equilibrium temperature reached! (assume no heat flows to the environment).

Answer:

Know: 𝑚₁ = 1 kg 𝑚₂ = 2 kg

𝑇₁= 80°

𝑇₂ = 30°

𝑐₁ = 4200𝐽

⁄𝐾𝑔℃

𝑐₂ = 900𝐽

⁄𝐾𝑔℃

Know: 𝑇_𝑚𝑖𝑥= ⋯ ? 𝑄_𝑜𝑢𝑡= 𝑄_𝑖𝑛

𝑚₁× 𝑐₁× 𝜃₁ = 𝑚₂× 𝑐₂× 𝜃₂ 1𝑘𝑔 × 4.200𝐽

⁄𝐾𝑔℃× (80℃ − 𝑇_𝑚) = 2𝑘𝑔 × 900𝐽

⁄𝐾𝑔℃× (𝑇_𝑚− 30℃)

4.200𝐽 ℃⁄

60 × (80℃ − 𝑇_𝑚) =^{1.800𝐽 ℃⁄}

60 × (𝑇_𝑚− 30℃)

70𝐽

⁄𝐾𝑔℃× (80℃ − 𝑇_𝑚) = 30 𝐽 𝐾𝑔℃⁄ × (𝑇_𝑚− 30℃)

5600𝐽 − 70𝐽

⁄ 𝑇℃ _𝑚= 30𝐽

⁄ 𝑇℃ _𝑚− 900𝐽 5600𝐽 + 900𝐽 = 70𝐽

⁄ 𝑇℃ _𝑚+ 30𝐽

⁄ 𝑇℃ _𝑚 6500𝐽 = 100𝐽

⁄ 𝑇℃ _𝑚

6500𝐽

100𝐽 ℃⁄ = ^{100𝐽 ℃⁄}

100𝐽 ℃⁄ 𝑇_𝑚 65℃ = 𝑇_𝑚

4. A total of 500 grams of water is heated from 20 ⁰C to 40 ⁰C. If the specific heat capacity of water is 4,200 J/kg⁰C, calculate the amount of heat received by the water (in joules).

Answer Answer:

Know: 𝑚 = 500 gr = 0,5 Kg 𝑇₁= 20℃

𝑇₂ = 40℃

θ = 𝑇₂− 𝑇₁ = 40℃ − 20℃ = 20℃

c = 4200𝐽

⁄𝐾𝑔℃

Unknow: 𝑄 = ⋯ ? 𝑄 = m × c × θ 𝑄 = 0,5 Kg × 4200𝐽

⁄𝐾𝑔℃× 20℃

Q = 42.000 J

5. If 5 kg of tea at 60 ⁰C is mixed into a glass filled with coffee at 8 kg at 5 ⁰C, what is the final temperature (T) of the mixture when equilibrium is reached, assuming no heat flows into the surroundings? (The specific heat of the water is 4200 J/kg ⁰C)

Answer:

Know: 𝑚₁ = 9 kg 𝑚₂ = 8 kg

𝑇₁= 60℃

𝑇₂ = 5℃

𝑐₁ = 𝑐₂ = 4200𝐽

⁄𝐾𝑔℃

Know: 𝑇_𝑚𝑖𝑥= ⋯ ? 𝑄_𝑜𝑢𝑡= 𝑄_𝑖𝑛

𝑚₁× 𝑐₁× 𝜃₁ = 𝑚₂× 𝑐₂× 𝜃₂ 9𝑘𝑔 × 4.200𝐽

⁄𝐾𝑔℃× (60℃ − 𝑇_𝑚) = 8𝑘𝑔 × 4.200𝐽

⁄𝐾𝑔℃× (𝑇_𝑚− 5℃)

9𝑘𝑔 ×^{4.200𝐽 ℃⁄}

4.200𝐽 ℃⁄ × (60℃ − 𝑇_𝑚) = 8𝑘𝑔 ×

4.200𝐽 ℃⁄

4.200𝐽 ℃⁄ × (𝑇_𝑚− 5℃)

9 × (60℃ − 𝑇_𝑚) = 8 × (𝑇_𝑚− 5℃) 540𝐽 − 9𝐽

⁄ 𝑇℃ _𝑚= 8𝐽

⁄ 𝑇℃ _𝑚− 40𝐽 540𝐽 + 40𝐽 = 8𝐽

⁄ 𝑇℃ _𝑚+ 9𝐽

⁄ 𝑇℃ _𝑚 540𝐽 = 17𝐽

⁄ 𝑇℃ _𝑚

540𝐽

17𝐽 ℃⁄ =^{17𝐽 ℃⁄}

17𝐽 ℃⁄ 𝑇_𝑚 31,76℃ = 𝑇_𝑚

6. The vaporization point of oxygen is 104 ⁰F. The vaporization point of oxygen when expressed in celsius is ….

Answer Answer:

Know: ℉ = 104℃

℃ = ⋯ ?

℉−IP

SP−IP= ^℃−IP

SP−IP

104−32°

212°−32°= ^℃−0°

100°−0°

72 180°= ^℃

100°

360℃ = 9℃

℃ = 40℃

7. The Y thermometer can measure ice point at -50⁰, and stem point on a scale of 150⁰. If the temperature of the object shows 20 ⁰C, then the temperature shown by the Y thermometer is ....⁰Y Answer

y−IP

SP−IP= ^℃−IP

SP−IP

y−(−50°)

150°−(−50)°=^20℃−0°_100°−0°

Y−50°

200° = ^20℃

y−50° 100°

2° = ^20℃

1°

y − 50° = 20℃

y = 70°