An aluminum calorimeter with a mass of 125 g contains 285 g of water. The calori

Question

An aluminum calorimeter with a mass of 125 g contains 285 g of water. The calorimeter and water are in thermal equilibrium at 20.0 degree C. A 75.0-g copper block at an initial temperature of 285 degree C is placed in the water. Assume no thermal energy is transferred to the environment. Calculate the equilibrium temperature of the calorimeter-water-block system. After the calorimeter-water-block system reaches thermal equilibrium, an unknown substance with a mass of 175 g and initial temperature of 95.0 degree C is placed in the water. If the entire system stabilizes at a final temperature of 29.6 degree C, calculate the specific heat of the unknown substance.

Explanation / Answer

here,

mass of alumunium , ma = 0.125 kg

mass of water , mw = 0.285 kg

mass of copper , mc = 0.075 kg

a)

let the equilibrium temrature be Tf

heat lost by copper = heat gained by water + heat gained by alumunium

mc * Cc * ( 285 - Tf) = mw * Cw * ( Tf - 20) + ma * Ca * (Tf - 20)

0.075 * 386 * ( 285 - Tf) = 0.285 * 4186 * ( Tf - 20) + 0.125 * 900 * ( Tf - 20)

solving for Tf

Tf = 25.7 degree

the final equilibrium temprature is 25.7 degree

b)

mass of unknown , mu = 0.175 kg

initial temprature , Ti = 95 degree C

Tf = 29.6 degree C

heat lost by unknown substance = heat gained by water + heat gained by alumunium

mc * Cc * ( 95 - Tf) = mw * Cw * ( Tf - 25.7) + ma * Ca * (Tf - 25.7)

0.175 * C * ( 95 - 29.6) = 0.285 * 4186 * ( 29.6 - 25.7) + 0.125 * 900 * ( 29.6 - 25.7)

C = 444.9 J/kg C

the specific heat of unknown substance is 444.9 J/kg C

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