A 19.6 g sample of ice at -10.0°C is mixed with 100.0 g of water at 75.4°C. Calculate the final temperature of the mixture assuming no heat loss to the surroundings. The heat capacities of H2O(s) and H2O(l) are 2.08 and 4.18 J g-1 °C-1, respectively, and the enthalpy of fusion for ice is 6.02 kJ/mol.

qice=-qliq

q=delta H ice * n
q=S(liq)*Mass(liq)*(Tf-Ti)?

Help please!

I try setting the two equal to each other and plug in the numbers but I still get it wrong.

(6020 J/mol)*(19.6g)*(1mol/18g)=(4.18 J g-1 C-1)(100g)(Tf-75.4)

To calculate the final temperature of the mixture, you can use the principle of conservation of energy. The heat gained by the colder object (ice) will be equal to the heat lost by the hotter object (water) until they reach thermal equilibrium.

Let's break down the steps to solve this problem:

1. Calculate the heat gained by the ice to reach 0°C:
- First, determine the heat required to raise the temperature of the ice from -10.0°C to 0°C.
Q_ice = mass_ice * specific_heat_ice * ΔT_ice
- Substituting the given values, we have:
Q_ice = (19.6 g) * (2.08 J g^(-1) °C^(-1)) * (0°C - (-10.0°C))

2. Calculate the heat gained by the water to reach 0°C:
- Since the water is initially at 75.4°C and needs to cool down to 0°C, we need to calculate the heat lost by the water.
- Q_water = mass_water * specific_heat_water * ΔT_water
(Note: The specific heat capacity of liquid water is used since it is transitioning from a liquid to solid phase)
- Substituting the given values, we have:
Q_water = (100.0 g) * (4.18 J g^(-1) °C^(-1)) * (75.4°C - 0°C)

3. Calculate the heat released during the phase change of the ice to water:
- The ice will melt and absorb heat during this process, but its temperature will remain at 0°C until it has completely melted.
- The heat required for this phase change is given by the enthalpy of fusion:
Q_fusion = mass_ice / molar_mass_ice * enthalpy_fusion
- Substituting the given values, we have:
Q_fusion = (19.6 g) / (18.0 g/mol) * (6.02 kJ/mol)

4. Calculate the final temperature of the mixture:
- Since no heat loss occurs, the heat gained by the ice should be equal to the heat lost by the water during cooling and phase change.
- Q_ice + Q_water + Q_fusion = 0
- (19.6 g) * (2.08 J g^(-1) °C^(-1)) * (0°C - (-10.0°C)) + (100.0 g) * (4.18 J g^(-1) °C^(-1)) * (75.4°C - 0°C) + (19.6 g) / (18.0 g/mol) * (6.02 kJ/mol) = 0

Now, solve this equation to find the final temperature of the mixture.