How much ice (at 0â°c) must be added to 1.90 kg of water at 79 â°c so as to end up with all liquid at 8 â°c? (ci = 2000 j/(kg.â°c), cw = 4186 j/(kg.â°c), lf= 3.35 ä‚— 10 5 j/kg, lv= 2.26 ä‚— 10 6 j/kg) kg

m = mass of the ice added = ?

M = mass of water = 1.90 kg

= specific heat of the water = 4186 J/(kg ⁰C)

 = specific heat of the ice = 2000 J/(kg ⁰C)

= latent heat of fusion of ice to water = 3.35 x 10⁵ J/kg

 = initial temperature of ice = 0 ⁰C

= initial temperature of water = 79 ⁰C

T = final equilibrium temperature = 8 ⁰C

using conservation of heat

Heat gained by ice = Heat lost by water

m  (T - ) + m  = M  ( - T)

inserting the values

m (4186) (8 - 0) + m (3.35 x 10⁵ ) = (1.90) (4186) (79 - 8)

m = 1.53 kg

the answer is false

explanation:

answer: at 24 february, 2014 the population gets twice of its initial population.

explanation:

since we have given that

the population for alpha city, t years after january 1, 2004 is given as

first we find out the initial population, i.e. t=0,

so, our quadratic equation becomes,

according to question, we have said that if the population has twice its initial population, then it becomes

so, time taken to reach this above mentioned population is given by

now, we know that time can't be negative so, we take t=9.149 years.

hence, at 24 february, 2014 the population gets twice of its initial population.