Apiece of metal with a mass of 1.9 kg, specific heat of 202 j/kg/◦c, and initial temperature of 83◦c is dropped into an insulated jar that contains liquid with a mass of 4 kg, specific heat of 1000 j/kg/◦c, and initial temperature of 7◦c. the piece of metal is removed after 2.7 s, at which time its temperature is 16◦c. find the temperature of the liquid after the metal is removed. neglect any effects of heat transfer to the air or to the insulated jar. answer in units of ◦c.

13.428° C

Explanation:

Given:

Mass of the metal piece, m = 1.9 kg

specific heat of the metal, C = 202 J/kg°C

Initial temperature of the metal, T₁ = 83° C

Final temperature of  the metal, T₂ = 16° C

mass of the liquid, M = 4 kg

specific heat of the liquid, C' = 1000 J/kg°C

Initial temperature of the liquid, T₃ = 7° C

Now,

when the metal is removed

The heat lost by the metal = The heat gained by the liquid

thus,

mCΔT = MC'ΔT'

let the final temperature of the liquid be T₄

thus, we have

- [1.9 × 202 × (T₂ - T₁)] = 4 × 1000 × (T₄ - T₃)     (here, negative sign depicts the heat lost by the metal)

on substituting the further values, we get

- [1.9 × 202 × (83 - 16)] = 4 × 1000 × (T₄ - 7)

or

T₄ - 7 =  6.42865

or

T₄ = 13.428° C

Hence, the final temperature of the liquid is 13.428° C

i think it would be 5

energy stored in the capacitor= 9.72 x 10⁻⁴ j

explanation:

the energy stored in the capacitor is given by

e= 1/2 c v²

e= energy stored

c= capacitance= 1.5 µc= 1.5 x 10⁻⁶f

v= potential difference=36 v

so energy stored= e= 1/2 (1.5 x 10⁻⁶) (36)²

energy stored= 9.72 x 10⁻⁴j