A 100 g copper calorimeter cup contains 235 cm3 of water and 45 g of ice, all in

Question

A 100 g copper calorimeter cup contains 235 cm3 of water and 45 g of ice, all in thermal equilibrium. A 240 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 30 C.

A)What is the mass of water before the hot iron is put into the cup? Express your answer using three significant figures??

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 30 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

a) Mass of the water = m_water = density of water * volume of the water

m_water = 1000 * (235*10^-6) = 0.235 kg = 235 gm

b) energy required to melt all of the ice is Q1 = m_ice*Lf

Lf is the latent heat of fusion = 333.55 J/g

Q1 = m_ice*Lf = 45*333.55 = 15009.75 J

C) Q2 = Q1+[(m_water+m_ice)*S_w*dT] + (m_cu*S_cu*dT)

Q2 = 15009.75+ ((235+45)*4.186*30) + (100*0.385*30)

Q2 = 5.13*10^4 J

D) by the principle of caloriemeter,heat lost by Iron = Q2 = 5.13*10^4 J

E) Q = 5.13*10^4 J

m_iron* S_iron*dT = 5.13*10^4

240*0.450*(T-30) = 5.13*10^4

T = 505 C

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