Answer:
The correct answer is "5217 Cal".
Explanation:
The given values are:
Specific heat,
c = 3.76 cal/g°C
Mass,
m = 25.0 g
Initial temperature,
T₁ = 21.5°C
Final temperature,
T₂ = 77.0°C
Now,
The heat energy will be:
⇒ [tex]Q=mc \Delta t[/tex]
On substituting the given values, we get
⇒ [tex]=25.0\times 3.76\times (77-21.5)[/tex]
⇒ [tex]=25\times 3.76\times 55.5[/tex]
⇒ [tex]=5217 \ Cal[/tex]
