Question

14. If 10 g of a metal at 80 degrees cause 100 g of water to heat by 3 degrees, what is the specific heat capacity of the metal?

15. How much energy it takes to melt 3.0 g of water (given the enthalpy of fusion of water)?

16. How much energy is released when 50 g of water are cooled from 70 to 20 degrees (given the specific heat capacity of water)?

Answer 1

14) Let the specific heat capacity of the metal be S; therefore, heat lost by the metal = (mass of the metal)*(specific heat capacity of the metal) = (10 g)*S = 10S g.

Heat gained by water = (mass of water)*(specific heat capacity of water)*(change in temperature) = (100 g)*(1 cal/g.°C)*(3 °C) = 300 cal (specific heat capacity of water = 1 cal/g.°C).

According to the principle of thermochemistry,

heat lost by the hot substance = heat gained by the cold substance

====> 10 S g = 300 cal
====> S = (300 cal)/(10 g) = 30 cal/g

The specific heat of the metal is 30 cal/g (ans).