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

metals are good conductor of electricity. it has free electrons on their surface. according to electron pool theory, each atom present in a metal crystal losses all of its valence electron and as a result a pool of free is form. these pool of electrons are free to move and   conduct electricity easily   while making metal as good conductor.

the correct answer is a. the ability of electrons to flow throughout the metal.

if it's not useful, feel free to report for mistake in answer.