Question

Calculate the specific heat of a substance of a 35g sample absorbs 48 J as the temperature is raised from 293 K to 313 K

Answer 1

Answer:

0.0686 J/g ° C

Explanation:

The specific heat capacity is the amount of heat needed to raise 1 gram of a substance by 1° C

We are given temperatures in Kelvin; so we need to convert it to ° C

° C = K - 273.15

For the initial temperature T₁ = 293 K

T₁ = 293 - 273.15  ° C

T₁ = 19.85  ° C

For the final temperature T₂

T₂ = 313 - 273.15  ° C

T₂ = 39.85  ° C

Change in temperature =  ΔT = T₂ - T₁

ΔT = 39.85 ° C - 19.85 ° C

ΔT = 20.00 ° C

mass given = 35 g

heat absorbed (q) = 48 J

specific heat (c) = ????

Using the expression:

q = mcΔT

we have:

48 = 35 × c × 20

48 = c  × 700

c = $\frac{48}{700}$

c = 0.0686 J/g ° C