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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?

Respuesta :

The specific of the unknown substance that underwent a change in temperature is 0.567 J/g-C. This was solved using the equation for heat transfer which is:

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.
Edufirst

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