Question

2.4. Calorimeter problem to find the specific heat of lead: If 110 g of lead shot at 120°C is dropped into 60 g of water at 25°C contained in a copper calorimeter of 45 g mass, what is the resulting temperature? [Given: Cwater = 1 cal/g/°C; Cpb = 0.031 cal/g/°C; Ccu = 0.093 cal/g/°C]

Answer 1

Heat lost by lead will be equal to heat gained by copper and water. We use this to find the required final temperature as shown below***********************************************************************************************
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Let the final temperature be T, Heat lost by lead is, By = mSAT - (1109) 0.031 callg/) (100°C - T) Heat gained by copper and water is, 82= mis, ata + MzS₂ A Tq = 459) 0.093 Calg/c)(T- 25°c)+(609){cal glºs) (T-25°C) = 64.185 (T-250) foom energy conservation Qy - B2 (1109) 0.031 Callg/°C)(100°C - 1) = 64.185 (T-250) T = 28.81°C