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Calculating Specific Heat Extra Practice Worksheet Key, Exercises of Physics

TSFX Australian university document for calculating specific heat worksheet with solutions key

Typology: Exercises

2020/2021

Uploaded on 04/20/2021

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Calculating Specific Heat
Extra Practice Worksheet
Q = mc∆T, where Q = heat energy, m = mass, and T = change in temp.
Remember, ∆T = (Tfinal Tinitial). Show all work and proper units.
1. A 15.75-g piece of iron absorbs 1086.75 joules of heat energy, and its temperature
changes from 25°C to 175°C. Calculate the specific heat capacity of iron.
2. How many joules of heat are needed to raise the temperature of 10.0 g of aluminum from
22°C to 55°C, if the specific heat of aluminum is 0.90 J/g°C?
3. Calculate the specific heat capacity of a piece of wood if 1500.0 g of the wood absorbs
67,500 joules of heat, and its temperature changes from 32°C to 57°C.
Q
m
C
ΔT
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Calculating Specific Heat

Extra Practice Worksheet

Q = mc∆T , where Q = heat energy , m = mass , and ∆T = change in temp. Remember, ∆T = (Tfinal – Tinitial). Show all work and proper units.

  1. A 15.75-g piece of iron absorbs 1086.75 joules of heat energy, and its temperature changes from 25°C to 175°C. Calculate the specific heat capacity of iron.
  2. How many joules of heat are needed to raise the temperature of 10.0 g of aluminum from 22°C to 55°C, if the specific heat of aluminum is 0.90 J/g°C?
  3. Calculate the specific heat capacity of a piece of wood if 1500.0 g of the wood absorbs 6 7,500 joules of heat, and its temperature changes from 32°C to 57°C.

Q

m C^ ΔT

  1. 100.0 g of 4.0°C water is heated until its temperature is 37°C. Calculate the amount of heat energy needed to cause this rise in temperature.
  2. 25.0 g of mercury is heated from 25°C to 155°C, and absorbs 455 joules of heat in the process. Calculate the specific heat capacity of mercury.
  3. What is the specific heat capacity of silver metal if 55.00 g of the metal absorbs 47.3J of heat and the temperature rises 15.0°C?

Answers

Q = mc∆T , where Q = heat energy , m = mass , and ∆T = change in temp. Remember, ∆T = (Tfinal – Tinitial). Show all work and proper units.

  1. A 15.75-g piece of iron absorbs 1086.75 joules of heat energy, and its temperature changes from 25°C to 175°C. Calculate the specific heat capacity of iron. C = Q = 1086.75 = 0.46 J/g°C m(Tf-Ti) 15.75(175-25)
  2. How many joules of heat are needed to raise the temperature of 10.0 g of aluminum from 22°C to 55°C, if the specific heat of aluminum is 0.90 J/g°C? Q = mC(Tf – Ti) = 10.0g (0.90J/g°C)(55-22) = 297 J
  3. Calculate the specific heat capacity of a piece of wood if 1500.0 g of the wood absorbs 6 7,500 joules of heat, and its temperature changes from 32°C to 57°C. C = Q = 67500 J = 1.8 J/g°C m(Tf-Ti) (1500 g)(57-32)
  4. 100.0 g of 4.0°C water is heated until its temperature is 37°C. Calculate the amount of heat energy needed to cause this rise in temperature. Q = mC(Tf-Ti) = 100g(4.184J/g°C)(37 – 4) = 14000 J
  5. 25.0 g of mercury is heated from 25°C to 155°C, and absorbs 455 joules of heat in the process. Calculate the specific heat capacity of mercury. C = Q = 455 J = 0.14 J/g°C m(Tf-Ti) (25g)(155-25)
  1. What is the specific heat capacity of silver metal if 55.00 g of the metal absorbs 47.3J of heat and the temperature rises 15.0°C? C = Q = 47.3 J = 0.0573 J/g°C m(Tf-Ti) (55.00g)(15)
  2. What mass of water will change its temperature by 3 0 C when 525 J of heat is added to it? m = Q = 525 J = 40 g C(Tf–Ti) (4.184J/g°C)(3)
  3. A 0.3 g piece of copper is heated and fashioned into a bracelet. The amount of energy transferred by heat to the copper is 66,300 J. If the specific heat of copper is 390 J/g 0 C, what is the change of the copper's temperature? ΔT = Q = 66,300 J = 600°C mC 0.3g(390J/g°C)