What is the change in temperature in a 128g sample of water if it absorbs 2808j of heat energy at a temperature of 2°c?

Given:

128g sample of water

2808J of heat energy

Required:

Change in temperature

Solution:

This can be solved through the equation H = mCpT where H is the heat, m is the mass, Cp is the specific heat and T is the change in temperature.

The specific heat of the water is 4.18 J/g-K

Plugging in the values into the equation

H = mCpT

2808J = (128g) (4.18 J/g-K) T

T = 5.25 K

n+h→nh3

we must remember that nitrogen and hydrogen are both diatomic molecules in their standard gas form. this adjusts the equation to

n2+h2→nh3

now we need to adjust coefficients in order to balance the atoms on each side of the equation. currently we have 2 atoms of nitrogen and 2 atoms of hydrogen on the reactant side and 1 atom of nitrogen and 3 atoms of hydrogen on the product side.

we can balance the hydrogens by placing a coefficient of 2 in front of ammonia and a coefficient of 3 in front of the hydrogen.

n2+3h2→2nh3

this gives us 6 hydrogen on each side and coincidentally the nitrogens now equal 2 on each side.