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2) In a calorimetry experiment, 70.0 g of a substance was heated to

270.00 degrees Celsius and then placed in 3500.0 g of water. The water
bath's starting temperature was 15.00 degrees Celsius. If the substance's
listed specific heat is 0.890 J/gC, predict the temperature at thermal
equilibrium. *
onlar

User Edbond
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1 Answer

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The temperature at thermal equilibrium : 16.1 °C

Further explanation

Q absorbed = Q released

Heat can be formulated :

Q = m.c.Δt

c= specific heat, J/g°C

Δt = temperature different, C°

m= mass, g

Q water=Q substance


\tt 3500* 4.18* (t_2-15)=70* 0.89* (270-t_2)\\\\14630(t_2-15)=62.3(270-t_2)\\\\14630t_2-219450=16821-62.3t_2\\\\14692.3t_2=236271\\\\t_2=16.1^oC

User Drk
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