A sample of an unknown substance has a mass of 0.158 kg. If 2,510.0 J of heat is required to heat the substance from 32.0°C to 61.0°C, what is the specific heat of the substance?
Answers
U = mCp(Tf-Ti)
where:
U = heat absorbed or released in joules
m = mass in grams
Cp = specific heat in J/g-C
Tf = final temperature
Ti = initial temperature
Substituting values and converting 0.158 kg to grams gives the value 0.567 J/g-C. You can use a table of specific heat values to determine what the substance is as well.
Answer: 0.469 J / g°C
Explanation:
1) Data:
m = 0.158 kg
Q = 2,150.0 J
ΔT = 61.0°C - 32.0°C
Cs = ?
2) Formula:
Q = m×Cs×ΔT
3) Solution
i) Solve for Cs: Cs = Q / [m×ΔT]
ii) Plug in the data and compute
Cs = 2,150.0 J / [158 g × (61.0°C - 32.0°C) ] = 0.469 J / g°C
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Answers
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To learn more about Chemical formula, refer to the below link:
# SPJ2
Answers
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