Question

Water (3110 g) is heated until it just begins to boil. If the water absorbs 5.01 x 10^5 J of heat in the process. What was the initial temperature of the water ? Express your answer with the appropriate units.

Answer 1

We use the equation;

Q = m(kg) x C(j/kg) x $\Delta$T(K)

Substituting gives

5.01 x 10⁵J = $\dfrac{3110}{1000}kgx4200J/kg x$ $\Delta$T

$\dfrac{5.01 x 10^5J}{3.11 x 4200J}=\Delta$T

$\Delta$T = 38.36K

Assuming that the water begins to boil at 100^oC,

Then initial temperature = Final temperature - $\Delta$T

= 100^oC - 38.36K = 61.64^oC