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Calculate what the specific heat capacity of an unknown metal is if 394)
of heat were released when 15.0 grams of the substance at 96.0°C was
cooled to 28.0°C?
Formula Reminders
q = mCAT
q = mHvap
q = mHfus
1. 4.01 J/gºC
2. 5910 J/g°C
3. 0.387 J/g°C

Calculate what the specific heat capacity of an unknown metal is if 394 of heat were released when 150 grams of the substance at 960C was cooled to 280C Formula class=

Respuesta :

Answer: The specific heat capacity of an unknown metal is [tex]0.387 J/g^{o}C[/tex].

Explanation:

Given: Heat energy = 394 J

Mass = 15 g

Initial temperature = [tex]96^{o}C[/tex]

Final temperature = [tex]28^{o}C[/tex]

Formula used is as follows.

[tex]q = m \times C \times (T_{2} - T_{1})[/tex]

where,

q = heat energy

m = mass of substance

C = specific heat capacity

[tex]T_{1}[/tex] = initial temperature

[tex]T_{2}[/tex] = final temperature

Substitute the values into above formula as follows.

[tex]q = m \times C \times (T_{2} - T_{1})\\394 J = 15 g \times C \times (28 - 96)^{o}C\\C = \frac{394 J}{15 g \times (68^{o}C)}\\= \frac{394 J}{1020 g^{o}C}\\= 0.387 J/g^{o}C[/tex]

Thus, we can conclude that the specific heat capacity of an unknown metal is [tex]0.386 J/g^{o}C[/tex].