An aluminum cup contains 150 g of water at 25 degree C. A 500 g silver sample at

Question

An aluminum cup contains 150 g of water at 25 degree C. A 500 g silver sample at an initial temperature of 100 degree C, is placed in the water. A 50 g copper stirrer is used to stir the mixture until it reaches its final equilibrium temperature of 35 degree C. Calculate the mass of the aluminum cup in grams. Assume the initial temperature of the stirrer is also 25 degree C (Assume the specific heat of aluminum, silver and copper are 0.215 cal/g. degree C, 0.056 cal/g. degree C and 0.0924 cal/g. degree C and that of water is 1 cal/g degree C)

Explanation / Answer

let,

mass of the aluminum cup = m_Al at T2=25 oC

mass of water mW=150g at T1=25 oC

mass of silver sample m_Ag=500g at T2=100 oC

and

mass of copper stirrer m_cu=50g at T1=25 oC

final equilibrium temperature is T_equ=35 oC

by using energy relation,

enegry lost = enegry gain

m_Ag*C_Ag*(T1-Tequ) = m_Al*C_Al*(T_equ-T2) + m_w*C_w*(T_equ-T2) + m_cu*C_cu*(Tequ-T2)

500*0.056*(100-35) = m_Al*0.215*(35-25) + 150*1*(35-25) + 50*0.0924*(35-25)

===> m_Al=127.35 g

mass of the Aluminum cup m_Al=127.35 g

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