Question

1.In the labratory a student finds that it takes 120 Joules to increase the temperature of 12.3 grams of solid sulfur from 22.9 to 37.3 degrees Celsius.

The specific heat of sulfur calcualted from her data is ___ J/g C.

2. the labratory a student finds that it takes 89.8 Joules to increase the temperature of 10.8 grams of solid graphite from 22.9 to 35.4 degrees Celsius.

The specific heat of sulfur calcualted from her data is ___ J/g C.

Answer 1

Answer:

Step 1: Explanation

We know

Q=m×c×ΔT

where, m = mass of the material, c =  specific heat capacity of the material, and ΔT = temperature change

(1) Step 2: Calculation of specific heat of solid sulfur

Mass of solid sulfur = 12.3 g

Initial temperature ( T₁ ) = 22.9 °C

Final temperature ( T₂ ) = 37.3 °C

ΔT = T_f − T_i = (37.3 -22.9 ) °C = 14.4 °C

Q = 120 J

On substituting the value in above equation

Q=m×c×ΔT

=> c = Q /m×ΔT

=> c = 120 J / (12.3 g × 14.4  °C ) = 0.6775 J/g °C

Hence, the specific heat of solid sulfur = 0.6775 J/g °C​​​​​​​

(2) Step 3: Calculation of specific heat of solid graphite

Mass of solid graphite = 10.8 g

Initial temperature ( T₁ ) = 22.9 °C

Final temperature ( T₂ ) = 35.4 °C

ΔT = T_f − T_i = (35.4 -22.9 ) °C = 12.5 °C

Q = 89.8 J

On substituting the value in above equation

Q=m×c×ΔT

=> c = Q /m×ΔT

=> c = 89.8 J / (10.8 g × 12.5  °C ) = 0.665  J/g °C​​​​​​​

Hence, the specific heat of solid graphite = 0.665 J/g °C​​​​​​​