Learning Goal: Calorimetry: Three Substances The ceramic coffee cup in the figure with m = 118 g and c = 1090 J/(kg - K), is initially at room temperature (24.0 °C). Coffee and cream will be added to the cup. Assume that no heat is exchanged with the surroundings, and that the specific heat of coffee and cream are the same as the specific heat of water, c = 4186 J/(kg.K) In both of the two situations, the same amount of coffee 220 g at the same initial temperature 79.0 C is added to the cup. But different amounts of cream of 6.0 °C is added to the cup. note: J/(kg. K) =J/(kg. °C) After 220 g of 79.0°C coffee and unknown amount of 6.0 °C cream are added to the cup, the equilibrium temperature of the system (Coffee + Cream + Cup) is measured to be 58.0 °C. How much cream in grams was added? Follow the steps in "hints". Enter the mass of the cream in Grams, keep 2 digits after the decimal point ► View Available Hint(s) mass of cream mcream = Submit 15| ΑΣΦ Part B www ? grams New Situation: The amount of coffee is still 220 g, its initial temperature is still 79.0 °C. Now the coffee and 23.0 g of 6.0 °C cream are added to the cup with the same mass and same initial temperature as in Part A. You will find the final temperature of the system (Coffee + Cream + Cup) when thermal equilibrium is reached. If 220 g of 79.0°C coffee and 23.0 g of 6.0 °C cream are added to the cup, what is the equilibrium temperature of the system? Follow the steps in "hints".