Question

Determine the number of grams of ice at 0∘C that must be added to a cup with 275 g of tea at 46 ∘C to cool the tea to 29 ∘C. The heat of fusion for water at 0 ∘C is Lf =3.35×105J/kg. The specific heat of water is c = 4180 J/kg⋅ ∘C.

Answer 1

To calculate the number of grams of ice needed to cool 275 g of tea from 46 °C to 29 °C, apply the principles of heat transfer and use the specific heat and heat of fusion for water to set the heat lost by the tea equal to the heat gained by the ice during melting and temperature rise to the final temperature.

Step-by-step explanation:

To determine the number of grams of ice at 0°C that must be added to 275 g of tea at 46 °C to cool the tea to 29 °C, we need to use the concept of heat transfer during the phase change and the specific heat capacity of water.

The energy lost by the tea as it cools down will be gained by the ice as it melts and comes to the same temperature as the tea.

We have the heat of fusion for water (Lf) as 3.35×105 J/kg, and the specific heat of water (c) as 4180 J/kg·°C.

The calculation involves setting the heat lost by the tea, which can be calculated using the specific heat formula (q = mcΔT), equal to the heat gained by the ice, which is the sum of the heat required to melt the ice and the heat required to raise the temperature of the melted ice to the final temperature, both calculated using specific heat and heat of fusion formulas accordingly.

First, calculate the heat lost by the tea using the formula:

1. qtea = mtea × cwater × (ΔTtea)
2. qtea = 275 g × 4180 J/kg·°C × (46 °C - 29 °C)

Next, calculate the heat required for melting the ice and bringing it up to the final temperature:

1. qice = mice × Lf + mice × cwater × (ΔTfinal)
2. Set qtea = qice and solve for mice

By solving these equations, we will find the mass of the ice required to achieve the desired temperature of the tea.