Calorimetry problem

Question

2.035g of a soluble salt is added to 48.1g of water in a calorimeter. The temperature changes from 26.6 °C to 19.4 °C. Calculate the heat change q of the reservoir in joules. Assume the heat capacity of the reservoir to be 4.184J/g°C.

Answer 1

q = mC∆T

q = heat = ?

m = mass = 2.035 g + 48.1 g = 50.135 g

C = specific heat = 4.184 J/g/deg

∆T = change in temperature = 26.6 - 19.4 = 7.2 degrees (note: temp. went DOWN meaning an endothermic rx)

q = (50.135 g)(4.184 J/g/deg)(7.2 deg) = 1510 J (positive sign since reaction is endothermic)