Answer: The correct answer is 1.43 cal.
Explanation:
The expression of heat in terms of mass and temperature is as follows;
[tex]Q= mc\Delta T[/tex]
Here, Q is the heat, m is the mass of the object and [tex]\Delta T[/tex] is the change in temperature.
It is given in the problem that 1.25 grams of silver is cooled from 100.0°C to 80.0°C.
Calculate the change in temperature.
[tex]\Delta T=100.0\°C- 80.0\°C[/tex]
[tex]\Delta T= 20\°C[/tex]
Calculate the heat in calories.
[tex]Q= mc\Delta T[/tex]
Put m= 1.25 grams, [tex]\Delta T= 20\°C[/tex] and c=0.057 cal/g°C.
[tex]Q= (1.25)(0.057)(20)[/tex]
Q= 1.43 cal
Therefore, the amount of heat is 1.43 cal.
