Question

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

Answer 1

The 13.9 g sample of CaCO3 absorbs approximately 132.01 joules of heat when the temperature increases from 21.7°C to 33.3°C, as calculated using the formula q = mcΔT with the given values.

Step-by-step explanation:

The amount of heat absorbed by a substance can be calculated using the specific heat capacity formula, which is q = mcΔT, where q is the heat absorbed, m is the mass of the sample, c is the specific heat capacity, and ΔT is the change in temperature. In this case, the temperature change (ΔT) is 33.3°C - 21.7°C, the mass (m) is 13.9 g, and the specific heat (c) is 0.82 J/g°K.

The calculation is done as follows:

ΔT = 33.3°C - 21.7°C = 11.6°K

q = m × c × ΔT

q = 13.9 g × 0.82 J/g°K × 11.6°K

q = 13.9 g × 0.82 J/g°K × 11.6°K = 132.01 J

Therefore, the sample of CaCO3 absorbs approximately 132.01 joules of heat when its temperature increases from 21.7°C to 33.3°C.

Answer 2

Q =132.22 j

Step-by-step 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