Question

Calculate the heat capacity, in joules and in calories per °C, of each of the following.

(a) 35.9 g of water at 35°C J/°C cal/°C

(b) 1.71 oz of selenium at 25°C J/°C cal/°C

Answer 1

Question a

Specific heat capacity of water = 4.814 J /g °C

Mass of water = 35.9 g

heat capacity of water = 35.9 g x 4.814 J /g °C = 150.21 J/ °C

Let us convet into calories  

1 cal = 4.184 J

150.21 J = 150.21 /4.184 = 35.9 cal

heat capacity of water is 150.21 J/ °C or  35.9 cal / °C

Question b

Specific heat capacity of Selenium = 0.32 J /g °C

Mass of Selenium = 1.71 Oz g

Let us convert into gram

1Oz = 28.35 gm

1.71 Oz = 28.35 x 1.71 =  48.478 gm

heat capacity of Selenium = 48.478 g x 0.32 J /g °C = 15.512 J/ °C

Let us convet into calories  

1 cal = 4.184 J

150.21 J = 15.512 /4.184 = 3.707 cal

heat capacity of Selenium is 15.512 J/ °C or   3.707​​​​​​​ cal / °C