Answer:
[tex]\fbox{c = - 4.01 \: joule/g°C}[/tex]
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:
[tex]Specific \: Heat \: c = \: ?[/tex]
Solution:
We know that,
Specific heat of any substance is directly proportional to the mass and change in temperature.
Represented by equation,
[tex]Q = mc \triangle T[/tex]
Where,
Q = Heat Energy
m = mass of given sample
c = specific heat
∆T = change in temperature
Substituting corresponding values,
[tex]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 \\ [/tex]
We also know that,
[tex]1 \: cal = 4.184 \: joules[/tex]
multiplying above answer by 4.184,
[tex]c = - 0.96 \times 4.184 \\ \fbox{c = - 4.01 \: joule/g°C}[/tex]
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