A person makes ice tea by adding ice to 1.8 kg of hot tea, initially at 80°C. How many kilograms of ice, initially at 0.00°C, are required to bring the mixture to 10°C? The heat of fusion of ice is 334 kJ/kg, and we can assume that tea has essentially the same thermal properties as water, so its specific heat is 4190 J/(kg K).

A mass of ice at -4°C is needed to cool 115 kg of vegetables in a bunker for 24 hours. The initial temperature of the vegetables is assumed to be 29°C. It is also assumed that, within the 24-hr period, the average temperature inside the bunker is 7°C. If the heat gained per hour in the bunker is 30% of the heat removed to cool the vegetables from 29°C to 7°C, what would be the required mass of ice in kg? Use: Specific heat of water 4.2292 kJ/kg-°C Specific heat of ice 1.9387 kJ/kg °C Specific heat of vegetables 3.35 kJ/kg-°C

You have 975 grams of liquid water in a 275 gram glass container. Initially, the water and the container both have a temperature of 22.5 ̊C. You would like to cool the water and the container to a final temperature of 10.5 ̊C. You have a large supply of ice, which is initially at 0 ̊C. How much ice would you need to add to the container in order to get the desired final temperature after the ice melts and thermal equilibrium is reached? Give your answer in grams. The specific heat of liquid water is 4,186 J kg−1 ̊C−1, the specific heat of glass is 845 J kg−1 ̊C−1, and the latent heat of fusion of water is 333,500 J kg−1. Assume that the glass + liquid water + ice is an isolated system (i.e., no heat flows in or out of the system during this process).