Question

Calculate the specific heat in J/(g·ºC) of an unknown substance if a 2.50-g sample releases 12.0 cal as its temperature changes from 25.0ºC to 20.0ºC. _________J/(g·°C)

Answer 1

Answer:

$\fbox{c = - 4.01 \: joule/g°C}$

Step by step explanation:

Given:

Mass of given sample (m) = 2.50 g

Initial temperature (T1) = 25°C

Final temperature (T2) = 20°C

Heat Energy Q = 12 cal

To find:

$Specific \: Heat \: c = \: ?$

Solution:

We know that,

Specific heat of any substance is directly proportional to the mass and change in temperature.

Represented by equation,

$Q = mc \triangle T$

Where,

Q = Heat Energy

m = mass of given sample

c = specific heat

∆T = change in temperature

Substituting corresponding values,

$Q = mc \triangle T \\ 12 = 2.5\times c \times (20-25) \\ c = \frac{12}{2.5 \times ( - 5)} \\ c = - 0.96 \: cal/g°C$

We also know that,

$1 \: cal = 4.184 \: joules$

multiplying above answer by 4.184,

$c = - 0.96 \times 4.184 \\ \fbox{c = - 4.01 \: joule/g°C}$

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