A 7.47 g sample of calcium carbonate [CaCO3 (s)] absorbs 85 J of heat, upon which the temperature of the sample increases from 21 °C to 34.9 °C. What is the specific heat of calcium carbonate? A 0.82
B 8800
C 46
D 160
E -0.82

Specific heat of calcium carbonate(C) = 0.82 (Approx)

Explanation:

Given:

Energy absorbs (q) = 85 J

Change in temperature (Δt) = 34.9 - 21 = 13.9°C  

Mass of calcium carbonate = 7.47 g

Find:

Specific heat of calcium carbonate(C)

Computation:

Specific heat of calcium carbonate(C) = q / m(Δt)

Specific heat of calcium carbonate(C) = 85 / (7.47)(13.9)

Specific heat of calcium carbonate(C) = 85 / 103.833

Specific heat of calcium carbonate(C) = 0.8186

Specific heat of calcium carbonate(C) = 0.82 (Approx)

the ideal gas law equation is as follows

pv = nrt

where p - pressure

v - volume

n - number of moles

r - universal gas constant

t - temperature

so if the pressure, volume and temperature are already known

we are left with n and r

since r is the universal gas constant that has a known fixed value then r too is known

so we are left with 'n'

once we know temperature volume and pressure

we can find the number of moles of gas present using the ideal gas law equation

this is not my own answer, this is an answer for the same question by the user blahblahmali. i do not claim ownership of this account.