2. An aluminium calorimeter with a mass of 35.0 g contains 295 g of water. The c

Question

2. An aluminium calorimeter with a mass of 35.0 g contains 295 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 295°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 system reaches thermal equilibrium, an unknown substance with a mass of 187 g and initial temperature of 95.0°C is placed in the water. If the entire system stabilizes at a final temperature of 29.8°C, calculate the specific heat of the unknown substance

Explanation / Answer

2) a) Using principle of caloriemetry

Heat lost by the copper block = Heat gained by the aluminium + water

m_cu*S_cu*dT1 = (m_Al*C_Al*dT2)+(m_w*S_w*dT3)

(75*0.385*(295-T)) = (35*0.900*(T-20)) + (295*4.186*(T-20))

solving for T ,we get

T = 26.13 deg C

b) Heat lost by unknown substance = Heat gained by copper + al + water

m_unknown*S_unknown*dT1 = (m_cu*S_cu*dT2)+(m_Al*S_Al*dT2)+(m_w*S_w*dT2)

(187*S_unknown*(95-29.8)) = (75*0.385*(29.8-26.13))+(35*0.900*(29.8-26.13))+(295*4.186*(29.8-26.13))

S_unknown = 0.389 J/g-K

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