Akettle of water is at 14.5°c. its temperature is then raised to 50.0°c by supplying it with 5,680 joules of heat. the specific heat capacity of water is 4.186 joules/gram degree celsius. what is the mass of water in the kettle? express your answer to three significant figures. the mass of the water in the kettle is grams.

- 38.2 g.

Solution:- The equation used for solving this type of calorimetry problems is:

where, q is the heat energy, m is mass, c is specific heat and delta T is the change in temperature.

Water temperature is increasing from 14.5 degree C to 50.0 degree C.

 = 35.5 degree C

q is given as 5680 J and specific heat value is .

The equation could be rearranged for m as:

Let's plug in the values in it:

m = 38.2 g

So, the mass of water in the kettle is 38.2 g.

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.

the answer would be b i believe