Answer:
Q =132.22 j
Explanation:
Given data:
Mass of CaCO₃ = 13.9 g
Initial temperature = 21.7 °C
Final temperature = 33.3 °C
Specific heat of CaCO₃ = 0.82 j/g.k
Amount of heat absorbed = ?
Solution:
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 = T2 - T1
ΔT = 33.3°C - 21.7°C
ΔT = 11.6°C
Q = m.c. ΔT
Q = 13.9 g. 0.82 j/g.k. 11.6°C
Q =132.22 j
