Answer:
The final temperature of the entire water mixture, once all the ice has melted, stands at 12.9°C. It’s crucial to understand that in a closed system where no heat is lost, the total heat exchanged is 0.
This implies that the temperature decreased as the ice transferred heat to the water, leading to its cooling.
Explanation:
To begin:
Q1 = Q representing the heat gained from melting ice.
Q2 = Q denoting the heat lost by the water to melt the ice.
Q1 + Q2 = 0
Given that the ice starts at 0 °C, we must first determine the energy required to completely melt it. If the ice had been at a lower temperature, we would have raised it to 0 °C using the formula:
Q = mass × specific heat × (ΔT)
and then, to perform the state transition by utilizing the latent heat of fusion.
The heat of fusion for water at 0 °C is roughly 334 joules per gram.
Thus, Q = Hf × mass.
Q1 = 334 J/g × 8.32 g = 2778.88 J
For the water, we use:
Q = mass × specific heat × (ΔT)
Q2 = 55g × 4180 J/kg·K (Tfinal - T initial)
Converting 55 g to kg yields 0.055kg to maintain consistent units.
Q2 = 0.055kg × 4180 J/kg·K (Tfinal (unknown) - 25°)
Note that the temperature in degrees Kelvin (K) is the same for specific heat but differentiates as K vs. °C.
25°C = 298K
Q2 = 0.055kg × 4180 J/kg·K (Tfinal - 298K)
The conclusion:
Q1 + Q2 = 0
334 J/g × 8.32 g + 0.055kg × 4180 J/kg·K (Tfinal - 298K)
2778.88 J + 229.9 J/K (Tfinal - 298 K) = 0
2778.88 J + 229.9 J/K × Tfinal - 68510.2 J = 0
229.9 J/K × Tfinal = 68510.2 J - 2778.88 J
Tfinal = 65731.4 J / 229.9 K/J
Tfinal = 285.9 K
Tfinal = 285.9 K - 273K = 12.9 °C
