The final temperature of the water is determined as 50.55 ⁰C.
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Final temperature of the water</h3>
The final temperature of the water is determined from the following calculations;
Q = mcΔθ
Δθ = Q/mc
where;
- Q is the amount of energy = 81 kcal = 338904 J
- c is specific heat capacity of water = 4,200 J/kgC
Δθ = 338904 /(3.5 x 4200)
Δθ = 23.05 °C
Final temperature = T₁ + Δθ
Final temperature = 27.5°C + 23.05 °C = 50.55 ⁰C.
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Answer:
Colloidal can not be separated through filtration.
Suspension can be separated through filtration.
Explanation:
Colloidal:
Colloidal consist of the particles having size between 1 - 1000 nm i.e, 0.001- 1μm. While the pore size of filter paper is 2μm. That's why we can not separate the colloidal through the filtration. However it can be separated through the ultra filtration. In ultra filtration the pore size is reduced by soaking the filter paper in gelatin and then in formaldehyde. This is only in case of when solid colloidal is present, if colloid is liquid , there is no solid particles present and ultra filtration can not be used in this case.
Suspension:
The particle size in suspension is greater than 1000 nm. The particles in suspension can be separated through the filtration. These particles are large enough and can be seen through naked eye.
Answer: 17.78g
Explanation:
Assume there is no heat exchange with the environment, then the amount of heat taken by the steel rod, Q(s), is equal to the amount of heat lost by the water, Q(w), but with opposite sign.
Q(s) = -Q(w)
Remember, Q = mc(ΔΦ)
Where Q = amount of heat
m = mass of steel
c = specific heat capacity of steel
ΔΦ = Initial temperature T1 - Final temperature T2
Q = mc(T1-T2)
Recall, Q(s) = -Q(w). Then,
m(s)*c(s)*(T1s - T2s) = - m(w)*c(w)*(T1w - T2w)
Substituting each values
Note: m(w) = volume of water*density = 75mL*1g/mL = 75g
m(s)*0.452*(21.5-2) = -75*4.18*(21.5-22)
m(s)*8.814 = 156.75
m(s) = 156.75/8.814
m(s) = 17.78g
Therefore, the mass of steel is 17.78g
