An aluminium calorimeter with a mass of 125 g contains 285 g of water. The calor

Question

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

Explanation / Answer

From the given question,

mass of calorimeter(m1) = 125g

mass of water(m2)=285g

Initial temperature of calorimeter(T1)=Initial temperature of water(T2)=20 degrees

mass of copper block(m3)=75g

Initial temperature of copper block(T3)=285 degrees

Let the final temperature be T

c1,c2 and c3 are specific heat of calorimeter, water and copper block respectively.

Heat lost by copper block = heat gained by calorimeter + water

m3 c3 (T3-T)=m1c1(T-T1) + m2c2(T-T2)

m3c3T3 - m3c3T = m1c1T - m1c1T1 + m2c2T - m2c2T2

T= (m3c3T3 + m1c1T1 + m2c2T2)/(m3c3+m1c1+m2c2)

=(75 c3 285 + 125 c1 20 + 285 c2 20) / ( 75 c3 + 125 c1 + 285 c2)

Let unknown substance has specific heat c4

Heat gained = heat lost

(m1c1 + m2c2 + m3c3 )(T-29.5)=m4c4(95-29.6)

c4= (125 c1 + 285 c2 + 75 c3)(T-29.5)/(175 x 65.4)

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