Question

Water (2190 g ) is heated until it just begins to boil. If the water absorbs 5.29×10 to the 5th power Jules of heat in the process, what was the initial temperature of the water? Specific heat capacity for water is 4.184 J/(g*c)

Answer 1

Answer – Given mass of water = 2190 g , heat, q = 5.29*10⁵ J , ti = ?

tf = 100.0^oC , specific heat of water = 4.184 J/g^oC

there is water is not completely boiled, so no need to heat of vaporization

q = m*C*∆t

5.29*10⁵ J = 2190 g * 4.184 J/g^oC*(100-ti)

5.29*10⁵ J = 916296 – 9162.96ti

9162.96ti = 916296 - 5.29*10⁵

               = 387296 J

So, ti = 387296 / 9162.96

         = 42.3^oC.