Answer:
Q = 96.6 j
Explanation:
Given data:
Heat required = ?
Initial temperature = 19°C
Final temperature = 33°C
Mass of disc = 3.0 g
Specific heat capacity = 2.3 J/g.°C
Solution:
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 33°C - 19°C
ΔT = 14°C
Q = 3.0 g×2.3 J/g.°C × 14°C
Q = 96.6 j
