Specific Heat Capacity?

Question

What is the final temperature of 100 g of water at 25 K is mixed with 40g of water at 90 K (SHC water 4200 J/kg K)?

Answer 1

I think you mean Celsius, not Kelvin, because you wouldn't be able to mix two blocks of ice.

The amount of heat given up by the warmer water will equal the amount of heat absorbed by the cooler water. Since the temperature difference for the warmer water will be negative, you have to have a negative on that side of the equation to cancel that out. As they are the same material, the specific heat will actually drop out of the equation.

-q(warm) = q(cool)
-(SH)*mw*deltaT = (SH)*mc*deltaT
-40*(x-90) = 100*(x-25)
-40x + 3600 = 100x - 2500
6100 = 140x
x = 43.57

Answer 2

I will assume you mean Degree Celsius as Kelvin seems tobe too cold. wE can write an energy and mass balance.

First the energy balance

Mwc*SHC*Tc +Mwh*SHC *Th= (Mwc+Mwh)*SHC*Tmix

where Mwc is mass of cold water, Mwh is mass of hot water SHC is the specific heat capacity, Tc is the temperature of cold water, Th is the temperature of hot water, Tmix is the final temperature of the mixture.

0.1*4200*(25+273) +0.04*4200*(90+273) = 0.14 *4200 *Tmix

125160 + 60984 = 588Tmix

186144=588Tmix

Tmix = 186144/588=316.57K= 43.42 Degree Celsius.

Answer 3

*water exists as ice at 25K and 90K
Nonetheless,

As heat travels from a hot to a cold medium, and as they both are mixed, they will equilibriate to the same final temperature(T)

By the conservation of energy,
heat given out by water at 90K (Q1)=heat absorbed by water at 25K(Q2)

Q1=(m)(c)(change in temp.)
=(40/1000)(4200)(90-T)

Q2=(m)(c)(change in temp.)
=(100/1000)(4200)(T-25)

equate Q1 and Q2 and solve for T(final temp.):
(40/1000)(4200)(90-T)=(100/1000)(4200)(T-25)
15120-168T=420T-10500
T=43.6K (final temperature)