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".
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".
Chapter1: Temperature And Heat
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Heat capacity is the amount of heat energy absorbed or released by a chemical substance per the change in temperature of that substance. The change in heat is also called enthalpy. The SI unit of heat capacity is Joules per Kelvin, which is (J K-1)
Thermal Properties of Matter
Thermal energy is described as one of the form of heat energy which flows from one body of higher temperature to the other with the lower temperature when these two bodies are placed in contact to each other. Heat is described as the form of energy which is transferred between the two systems or in between the systems and their surrounding by the virtue of difference in temperature. Calorimetry is that branch of science which helps in measuring the changes which are taking place in the heat energy of a given body.
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Transcribed Image Text: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".
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