1. A 120 g copper calorimeter cup contains 210 cm3 of water and 25 g of ice, all

Question

1. A 120 g copper calorimeter cup contains 210 cm3 of water and 25 g of ice, all in thermal equilibrium. A 220 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 is 26 C.

Part A

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

Express your answer using three significant figures.

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Part B

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

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Part C

How much energy is required to raise the temperature of the melted ice and water and copper cup to 26 C? (Signs matter!)

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Part D

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

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Part E

What was the initial temperature of the iron?

2.

Part A

To what temperature do you need to heat both the brass cylinder and the aluminum ring so that the ring fits over the cylinder?

Express your answer using three significant figures.

1. A 120 g copper calorimeter cup contains 210 cm3 of water and 25 g of ice, all in thermal equilibrium. A 220 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 is 26 C.

Part A

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

Express your answer using three significant figures.

Mass of Water = g

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Part B

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

Energy = kJ

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Part C

How much energy is required to raise the temperature of the melted ice and water and copper cup to 26 C? (Signs matter!)

Energy = kJ

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Part D

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

Energy = kJ

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Part E

What was the initial temperature of the iron?

Temperature = C

Explanation / Answer

(A) V = 210 cm^3 of water

m_water = 210 g

(B) Q = m Lf

Q = 25 x 333.55 = 8338.75 J Or 8.34 kJ

(C) Q = m c deltaT

Q = [ (25 + 210) (4.186 ) (26) ] + [ 120 x 0.385 x 26]

= 26777.7 J Or 26.8 kJ

(D) from energy conservatiom,

Q = 26.8 kJ

(E) specific heat capacity of iron = 0.450 J / g deg C

26777.7 = 220 x 0.450 x (T - 26)

T - 26 = 296.5 deg C

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