Question

In the laboratory a student finds that it takes 183 Joules to increase the temperature of 13.0 grams of solid magnesium from 24.6 to 39.4 degrees Celsius. i The specific heat of magnesium calculated from her data is J/gºC. In the laboratory a student finds that it takes 595 Joules to increase the temperature of 11.2 grams of liquid water from 21.6 to 35.3 degrees Celsius. The specific heat of water calculated from her data is J/gºC.

Answer 1

Answer:

Step 1: Explanation

When the temperature of n moles of a substance, having specific heat at constant pressure is Cp is changed by ΔT, then the amount of heat transferred by the body is given by

Q = mcΔT

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 magnesium

Mass of substance = 13 g

Initial temperature ( T1 ) = 24.6 °C

Final temperature ( T2 ) = 39.4 °C

ΔT = Tf − Ti  = (39.4 -24.6 ) °C = 14.8 °C

Q = 183 J

On substituting the value in above equation

Q=m×c×ΔT

=> c = Q /m×ΔT

=> c = 183 J / (13 g × 14.8  °C ) = 0.95 J/g °C

Hence, the specific heat of magnesium = 0.95 J/g °C​​​​​​​

(2)

Step 3: Calculation of specific heat of water

Mass of substance = 11.2 g

Initial temperature ( T1 ) = 21.6 °C

Final temperature ( T2 ) = 35.3 °C

ΔT = Tf − Ti  = (35.3 -21.6 ) °C = 13.7 °C

Q = 595 J

On substituting the value in above equation

Q=m×c×ΔT

=> c = Q /m×ΔT

=> c = 595 J / (11.2 g × 13.7 °C ) = 3.88 J/g °C​​​​​​​

Hence, the specific heat of water = 3.88 J/g °C​​​​​​​