How many joules of heat are absorbed when the tempurature of a 13.9 g sample of caco3 (s) increases from 21.7c to 33.3 c? specific heat of calcium carbonate is .82 j/g-k

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

i know insulin is primary but i dont know the others im sorry

explanation: