Question

You wish to chill your freshly brewed tea with the minimum amount of ice that will avoid watering it down too much. What is the minimum amount of ice you should add to2.70 kgof freshly brewed tea at91.98∘Cto cool it to6.19∘C? The ice is initially at a temperature of -51∘C. Heat capacity of water a 4190 Jikgik The specifio heat of ice=2010 J/kg/KLatent heat coeficient of fusion for ice=3.34×105 J/kg

Answer 1

To find the minimum amount of ice needed to cool the tea, you can use the formula:

Q = mcΔT + mL

where Q is the heat energy required, m is the mass of the ice, c is the specific heat capacity of ice, ΔT is the temperature change, and L is the latent heat of fusion.

The specific heat capacity of ice is 2010 J/kg°C. The latent heat of fusion of ice is 3.34 x 10^5 J/kg.

First, find the heat energy needed to cool the tea from 91.98°C to 6.19°C:

Q = 2.70 kg * 4190 J/kg°C * (91.98°C - 6.19°C) = 2.70 kg * 4190 J/kg°C * 85.79°C = 92959.58 J

Then, find the heat energy needed to melt the ice:

Q = mL = m * 3.34 x 10^5 J/kg = m * 3.34 x 10^5 J/kg

Then, add the heat energy needed to cool the tea to the heat energy needed to melt the ice:

Q = 92959.58 J + mL = 92959.58 J + 3.34 x 10^5 J/kg * m

Finally, rearrange the equation to solve for m:

m = (Q - 92959.58 J) / (3.34 x 10^5 J/kg)

Plugging in the given values, you get:

m = (92959.58 J - 92959.58 J) / (3.34 x 10^5 J/kg) = 0 kg

So the minimum amount of ice needed to cool the tea is 0 kg. This means that the tea is already at a temperature lower than the melting point of ice, so no ice is needed to cool it further. It's important to note that this calculation assumes that there is no heat loss to the surroundings and that the tea and ice are perfectly insulated. In practice, some heat will be lost to the surroundings and more ice may be needed to achieve the desired temperature.