How many joules are absorbed when 86.8 grams of water is heated from 40.0 °c to 86.0 °c?
11,400 joules
12,600 joules
14,100 joules
16,700 joules

16700Joules

Explanation:

Given parameters:

Mass of water = 86.8grams = 0.0868Kg

Initial temperature = 40°C, to kelvin, 40+273 = 313K

Final temperature = 86°C to kelvin, 86 + 273 = 359k

Solution:

This problem is to find the heat quantity Q that would be required to raise the temperature of the water through the given temperature.

     Q = mc(T₂ - T₁)

where m is the mass of the water

          c is the specific heat capacity of water = 4200JKg⁻¹K⁻¹

         T₂ and T₁ are the final and initial temperatures respectively

Q = 0.086 x 4200 x (359-313) = 16,615.2Joules or 16700Joules