6. A serving of Cheez-Its releases 130 kcal (1 kcal = 4.18 kJ) when digested by your body. If this same amount of energy were transferred to 2.5 kg of water at 27°C, what would the final temperature be?

78.8°C

Explanation:

Given parameters:

Amount of heat energy  = 130kCal

Mass of the cheese  = 2.5kg

Initial temperature  = 27°C

Unknown:

Final temperature  = ?

Solution:

To solve this problem, use the expression below;

       H = m C (T₂ - T₁)

H is the amount of heat

m is the mass

C is the specific heat capacity of water  = 4200J/kg°C

Now,

     convert the amount of heat to kJ;

            1kCal  = 4.18KJ

           130kCal  = 130 x 4.18  = 543.4kJ

Insert the parameters and solve;

       543.4 x 10³ = 2.5 x 4200 x (T₂ - 27 )

         543.4 x 10³  = 10500 x  (T₂ - 27 )

           (T₂ - 27 ) = 51.75

            T₂ = 51.75 + 27  = 78.8°C