If 5430 j of energy is used to heat 1.25 l of room temp. water (23.0 °c) whats the final temp of the water?

We can use the heat equation,
Q = mcΔT 

 

Where Q is the amount of energy transferred (J), m is the mass of the substance (kg), c is the specific heat (J g⁻¹ °C⁻¹) and ΔT is the temperature difference (°C).

Density = mass / volume

The density of water = 0.997 g/mL

Hence mass of 1.25 L (1250 mL) of water = 0.997 g/mL x 1250 mL

                                                                   = 1246.25 g

Specific heat capacity of water = 4.186 J/ g °C.

Let's assume that there is no heat loss to the surrounding and the final temperature is T.

By applying the equation,

      5430 J = 1246.25 g x 4.186 J/ g °C x (T - 23) °C
(T - 23) °C = 5430 J / 1246.25 g x 4.186 J/ g °C
(T - 23) °C = 1.04 °C
               T = 1.04 °C + 23 °C
               T = 24.04 °C

Hence, the final temperature of the water is 24.04 °C.

Answerthis solution will not form a precipitateexplanationsolubility equilbrium:         pbf₂(s)  ⇄ pb²⁺(aq) + 2 f⁻(aq)the solubility product expression is    (we exclude solids from the equilibrium expression).now suppose we have the trial ion product q, which we can define as    then, we can interpret  q as the product of the concentration of the ions that we have and ksp as the product of the ion concentrations that are needed to form a saturated solution.if q > ksp, then that implies that there will be a surplus of ions that exceed the amount that is required to form a saturated solution. a precipitate will then form (the precipitate continually  forms until the surplus ions are removed and q is decreased to ksp's value).if q = ksp, the saturated solution that forms is barely saturated and the precipitate will be minimal.if q < ksp, then the ion concentration will not be enough to form a saturated solution (so no precipitate).using our information, we calculate the trial ion product:     since q = 2.7  × 10⁻⁸ < ksp =  4  ×  10⁻⁸, a precipitate will not form.