Final answer:
The energy required to melt 25 g of ice and increase the water temperature to 3.0 ºC is 8,815 J, which involves the latent heat of fusion and the specific heat capacity of water.
Step-by-step explanation:
A busy housewife left 25 g of ice in an open insulated container, which then transformed into water at 3.0 ºC from its initial state of 0 ºC. To calculate the energy needed for the ice to totally melt and reach this temperature, we consider two processes: the melting of ice and the heating of the resulting water.
The specific latent heat of fusion of ice (α) is 340,000 J/kg, and the specific heat capacity of water (β) is 4,200 J/kgK.
To melt the ice, the energy required (Q1) is given by:
Q1 = m×α
where 'm' is the mass of the ice.
For the 25 g (0.025 kg) of ice:
Q1 = 0.025 kg × 340,000 J/kg = 8,500 J
After melting, the ice becomes water at 0 ºC. Now, to increase the temperature to 3.0 ºC, we use the equation:
Q2 = m×β×ΔT
where ΔT is the change in temperature. Q2 = 0.025 kg × 4,200 J/kgK × 3 K = 315 J
The total energy required is the sum of Q1 and Q2:
Total energy (Q) = Q1 + Q2 = 8,500 J + 315 J = 8,815 J
Hence, the energy needed to melt 25 g of ice and then increase the temperature of the water to 3.0 ºC is 8,815 J.\