a 6 0 kg piece of aluminum at 645 degrees to celsius is combine with 1.0 kg of ice at -15 degrees Celsius in a thermally isolated calorimeter of negligible heat capacity what is the equalilbrum temperature the mixture evenually reaches?.
q1+q2+q3=0
QAl = .6kg* .9kj/kg-K * 918K
4.96 KJ
To calculate the equilibrium temperature that the mixture eventually reaches, we can use the principle of conservation of energy. The equation Q1 + Q2 + Q3 = 0 represents the total energy exchange between the aluminum (Q1), ice (Q2), and the surroundings (Q3), where Q1, Q2, and Q3 represent the heat absorbed or released by each component.
In this case, since the calorimeter is thermally isolated (negligible heat capacity), Q3 is equal to zero because there is no exchange of heat with the surroundings.
Now let's calculate the values of Q1 and Q2:
Q1 (heat absorbed or released by aluminum):
Mass (m1) = 6.0 kg
Specific heat capacity (c1) = 0.9 kJ/kg-K (assuming specific heat capacity of aluminum)
Change in temperature (ΔT1) = final temperature - initial temperature = Tf - 645°C
To convert the temperature to Kelvin, we add 273.15:
Tf (final temperature in Kelvin) = (Tf in °C) + 273.15
Q1 = m1 * c1 * ΔT1
Q2 (heat absorbed or released by ice):
Mass (m2) = 1.0 kg
Specific heat capacity (c2) = 2.09 kJ/kg-K (assuming specific heat capacity of ice)
Change in temperature (ΔT2) = final temperature - initial temperature = Tf - (-15°C)
To convert the temperature to Kelvin, we add 273.15:
Tf = (Tf in °C) + 273.15
Q2 = m2 * c2 * ΔT2
Since Q1 + Q2 + Q3 = 0, we can solve for Tf:
Q1 + Q2 + Q3 = 0
[ m1 * c1 * (Tf - 645) ] + [ m2 * c2 * (Tf - (-15)) ] = 0
Substituting the given values:
[6.0 kg * 0.9 kJ/kg-K * (Tf - 645)] + [1.0 kg * 2.09 kJ/kg-K * (Tf - (-15))] = 0
Now, solve this equation for Tf to find the equilibrium temperature the mixture eventually reaches.