A 26.0 g sample of ice at -17.0°C is mixed with 124.0 g of water at 81.0°C. What is the equilibrium temperature?ºC.
heat gained by ice moving from -17 to zero + heat gained by ice melting |+ heat lost by warm water + heat gained by cool water = 0
[mass ice x specific heat solid ice x (Tfinal-Tinitial) + (mass ice x heat fusion at zero C) + [masswarm H2O x specific heat H2O x (Tfinal-Tintial)] |mass cool H2O x specific heat H2O x (Tfinal-Tinitial)] = 0
Solve for Tfinal.
To find the equilibrium temperature of the mixture, you can apply the principle of conservation of energy.
The equation for conservation of energy is:
m1 * c1 * ΔT1 + m2 * c2 * ΔT2 = 0
Where:
m1 = mass of ice
c1 = specific heat capacity of ice
ΔT1 = change in temperature of ice
m2 = mass of water
c2 = specific heat capacity of water
ΔT2 = change in temperature of water
In this case, the ice is initially at -17.0°C, and the water is initially at 81.0°C. We need to find the equilibrium temperature, which means ΔT1 and ΔT2 will have the same value. Let's assume the equilibrium temperature is T.
Using the equation, we can rearrange it to solve for T:
m1 * c1 * (T - (-17.0)) + m2 * c2 * (T - 81.0) = 0
Now, let's plug in the values we have:
m1 = 26.0 g
c1 = specific heat capacity of ice (2.09 J/g°C)
m2 = 124.0 g
c2 = specific heat capacity of water (4.18 J/g°C)
The equation becomes:
26.0 * 2.09 * (T - (-17.0)) + 124.0 * 4.18 * (T - 81.0) = 0
Simplifying the equation will give you a quadratic equation in terms of T. By solving the quadratic equation, you can find the value of T, which will be the equilibrium temperature of the mixture.