How Does Mixing Coffee, Cream, and a Ceramic Cup Affect Temperature Equilibrium?

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The discussion focuses on calculating the equilibrium temperature when mixing coffee, cream, and a ceramic cup, assuming no heat loss to the environment. The specific heat of coffee and cream is considered equivalent to that of water for the calculations. Participants suggest using the energy balance equation, which equates the heat gained by the cream and cup to the heat lost by the coffee. Clarification is provided on how to properly set up the equation to solve for the equilibrium temperature. The conversation emphasizes understanding the relationship between heat transfer and temperature changes in the system.
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A ceramic coffee cup, with m = 137 g and c = 1090 J/(kgK), is initially at room temperature (24.0°C). If 265 g of 80.3°C coffee and 12.2 g of 5.00°C cream are added to the cup, what is the equilibrium temperature of the system? 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.

I tried to use the equation c=Q/(mdeltaT) but I'm not sure what to plug in for Q because I can't calculate its value. Am I completely off track about the equation?
 
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Ummm... Sort of. Try writing out the whole equation that relates increase in energy and decrease in energy.
 
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