Answer:
6.82 kg
Explanation:
Given that the amount of water is 15L and we know that the density of water is ≈ 1kg/L. The mass of water is given by mass = volume x density, i.e,
mass = 15 x 1 = 15 kg. Also the specific heat capacity of water is 4.186 KJ/kg.
The sublimation enthalpy of dry ice is 571 KJ/kg.
Now, the amount of heat lost by water is entirely used up for the sublimation (conversion from soild to gas) of dry ice. And the heat (Q) lost by water is given as : Q = mCΔT, where m is the mass of water, C the specific heat capacity of water and ΔT the change in temperature.
Here, Q = 15 x 4.186 x (90 - 28) = 3892.98 KJ.
This amount of heat is taken up by the dry ice for its sublimation. Also the energy taken by dry ice (Q') for its sublimation is given by: Q' = m'L', where m' is the mass of dry ice, L' is the latent heat of sublimation (i.e, the amount of heat required per kg of a substance to sublime) of dry ice amd L' = 571 KJ/kg.
Now, Q' =m'L' = heat lost by water = 3892.98KJ.
And, m'L' = m' x 571 KJ/kg = 3892.98 KJ. (Dividing with 571)
Therefore, m' = 6.82 kg.
Answer:
b
Explanation:
An acid-base titration is an experimental procedure used to determined the unknown concentration of an acid or base by precisely neutralizing it with an acid or base of known concentration. ... It is filled with a solution of strong acid (or base) of known concentration.
The specific heat capacity the substance is calculated using the below formula
Q(heat) = Mc delta T
Q =1560 cal
m(mass) 312 g
delta T (change in temperature ) = 15 c
C= specific heat capacity=?
by making c the subject of the formula
c=Q/m delta T
= 1560 cal/ 312g x 15 c = 0.33 cal/g/c (answer B)
