Suppose you are making yourself 0.50 liters of tea but its temperature is 75°C, which is too hot for you to drink. In order to cool its temperature, you immerse three ice cubes of the same dimension as the one in part b) in your tea. If the energy used to melt the ice came solely from the internal energy (i.e. temperature) of the tea, what would the reduced temperature of your tea be (where you may assume tea is made up entirely of water)? For simplicity, in this class you may generally use a value of 1000 kg/m3 for the density of water (although density, like many of the properties of water used in this problem have a relatively small temperature dependence).

2) the first law of thermodynamics which states that the energy that enters a system is the same as the one that goes out, so the energy of the hot tea plus the energy of the ice is equal to the energy of the mixture between the two.

3.Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)

through prior knowledge of two other properties such as pressure and temperature.

for ice

first we found the mass

[tex]density = \frac{m}{V} \\Mice=(density)(V)\\Mice=(1000)(0.000024)=0.024Kg\\\\[/tex]

using thermodynamic tables we find the internal energy with T = 0C and pressure = 1atm

Uice=0KJ/kg

for tea

first we found the mass

V=0.5L=0.0005M^3

[tex]Mtea=(1000)(0.0005)=0.5kg[/tex]

using thermodynamic tables we find the internal energy with T = 75C and pressure = 1atm

Utea=313.9KJ/kg

Mass of mixture

Mm=Mtea+Mice

Mm=0.5kg+0.024kg

now we use the first law of thermodynamics

Now we use the first law of thermodynamics to find the internal energy of the mixture

(Uice)(Mice)+(Utea)(Mtea)=MmUm

[tex]\frac{(Uice)(Mice)+(Utea)(Mtea)}{Mm} =Um\\\frac{(0)(0.024)+(313.9)(0.5)}{0.524} =Um\\Um=299.5KJ/kg[/tex]

finally we find the new temperature using internal energy and pressure using thermodynamic tables

T=71C