Specific Heat and Enthalpy

Question

You put 5g of ice into 30g liquid water at 20degrees Celcius. The water cools down to 5 degrees Celcius. What temperature was the ice?

Answer 1

To solve this problem, we'll use the conservation of energy principle. The heat gained by the ice must equal the heat lost by the liquid water. First, we will find the heat lost by the liquid water, then we will use that to find the initial temperature of the ice.

1. Calculate the heat lost by the liquid water: Q_water = mass_water * specific_heat_water * (T_initial - T_final) Q_water = 30g * 4.18 J/(g°C) * (20°C - 5°C) Q_water = 30g * 4.18 J/(g°C) * 15°C Q_water = 1881 J
2. Calculate the heat gained by the ice:

a) Heat to raise the ice temperature to 0°C: Q1_ice = mass_ice * specific_heat_ice * (T_final - T_initial) Q1_ice = 5g * 2.09 J/(g°C) * (0°C - T_initial)

b) Heat to melt the ice: Q2_ice = mass_ice * enthalpy_of_fusion_ice Q2_ice = 5g * 333.5 J/g Q2_ice = 1667.5 J

c) Heat to raise the melted ice (now water) temperature to 5°C: Q3_ice = mass_ice * specific_heat_water * (T_final - T_initial) Q3_ice = 5g * 4.18 J/(g°C) * (5°C - 0°C) Q3_ice = 104.5 J

Total heat gained by the ice: Q_ice = Q1_ice + Q2_ice + Q3_ice

Since the heat gained by the ice is equal to the heat lost by the liquid water: Q1_ice + Q2_ice + Q3_ice = Q_water Q1_ice = Q_water - Q2_ice - Q3_ice Q1_ice = 1881 J - 1667.5 J - 104.5 J Q1_ice = 109 J

Now, we can solve for the initial temperature of the ice:

Q1_ice = 5g * 2.09 J/(g°C) * (0°C - T_initial) 109 J = 5g * 2.09 J/(g°C) * (-T_initial) 109 J = 10.45 J/°C * (-T_initial)

T_initial = - (109 J / 10.45 J/°C) T_initial ≈ -10.4 °C

So, the initial temperature of the ice was approximately -10.4°C.