A 120 g copper calorimeter cup contains 185 cm3 of water and 25 g of ice, all in

Question

A 120 g copper calorimeter cup contains 185 cm3 of water and 25 g of ice, all in thermal equilibrium. A 195 g piece of iron is heated to a high temperature and dropped into the water (assume no water escapes the calorimeter cup). At thermal equilibrium the temperature of the entire mixture is24 C.

A. What is the mass of water before the hot iron is put into the cup?

B. How much energy is required to melt all of the ice? (Signs matter!)

C. How much energy is required to raise the temperature of the melted ice and water and copper cup to 24 C? (Signs matter!)

D. How much energy leaves the iron in the form of heat? (Signs Matter!)

E. What was the initial temperature of the iron?

Explanation / Answer

Part A)

p = m/V

1000 = m/1.85 X 10^-4

m = .185 kg

Part B)

Q = mH

Q = (.025)(3.34 X 10^5) = 8350 J

Part C)
specific heat of water - 4186

specific heat of copper - 387

Q = mc(delta T) + mc(delta T) + mH

Q = (.185 + .025)(4186)(24) + (.120)(387)(24) + 8350

Q = 30562 J

Part D)
The iron must lose the same amout of heat gained in part C

So the iron loses 30562 J (you may need a negative here since it is heat lost)

Part E)
specific heat of iron - 448

Q = mc(delta T)

30562 = (.195)(448)(delta T)

Delta T = 349.8 degrees C

Since it got down to 24. T = 349.8 + 24 = 373.8 degrees C

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