Find the final equilibrium temperature when 10.0 g of milk at 10.0°c is added to 160 g of coffee at 90.0°c.
(assume the specific heats of coffee and milk are the same as water and neglect the heat capacity of the container.) cwater = 1.00 cal/g·°c = 4186 j/kg·°c
a. 85.3°c
b. 77.7°c
c. 71.4°c
d. 66.7°c

The final equilibrium temperature is T = 85.3 °C.

Explanation:

Knowledge of specific heats and/or heat capacities and the fact that energy must be conserved allows us to determine the equilibrium temperature of two objects initially at different temperatures by demanding that,

                Heat lost by hot object +  Heat gained by cold object = 0

where we ignore heat gained or lost from/to the surroundings.

We can calculate the heat released or absorbed using the specific heat capacity C, the mass of the substance m, and the change in temperature ΔT in the equation:

where

Use the equation for heat transfer to express the heat lost by the coffee

Express the heat gained by the milk

Use the law of conservation of energy and solve for

                      Heat lost by hot object + Heat gained by cold object =0

The final equilibrium temperature is T = 85.3 °C.

  1.the balanced   chemical equation is as below

cu +   2 ag(no)₃ → cu(no₃)₂   +2 ag

2.the   much silver   produced   is   50.9 g

step 1: find the   moles of cu

moles =   mass÷   molar mass

  from periodic table the molar mass of cu = 63.5 g/mol

moles is therefore = 15.00 g÷ 63.5 g/mol = 0.236   moles

step 2: use the mole ratio to determine the   moles of ag

from equation above   cu: ag   is 1: 2

therefore the moles of ag   =0.236   moles   x 2/1 =0.472   moles

step 3: find the mass   of ag

mass = moles x molar mass

from periodic table   the molar mass of ag =107.87 g/mol

mass   = 0.472   moles x 107.87 g/mol = 50.9 g

3.   the   much   silver   collected worth is   1.797 oz

  convert   50.9 g   to oz

that is   1 gram = 0.0353 oz

          50.9 grams =?   oz

by cross multiplication

={(50.9 grams x 0.0353 oz) / 1   gram}   =1.797 oz

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