Question

50.0 grams of cold water is placed in a styrofoam calorimeter and the initial temperature of this water is 22.0 C. then 50.0 grams of warm water, initially at 40.0 C, is added to the cold water and the final temperature of the mixture is 30.6 C. Calculate: a) heat absorbed by cold water; heat lost by warm water, anc c) heat capacity of calorimeter

Answer 1

Answer – We are given, mass of cold water = 50.0 g, cold water temp, ti = 22.0 ^oC , mass of warm water = 50.0 g , initial temp of warm water = 40.0 ^oC

Final temp, tf = 30.6^oC

a)Heat absorbed by cold water

we know heat formula

q = m x C x ∆T

= 50.0 g x 4.184 J/^oC.g x (30.6^o – 22.0^oC)

= 1799.12 J

So, heat is absorbed by cold water is 1799.12 J

b) Heat absorbed by warm water

we know heat formula

q = m x C x ∆T

= 50.0 g x 4.184 J/^oC.g x (30.6^o – 40.0^oC)

= -1966.48 J

So, heat lost by warm water is -1966.48 J

c) Heat capacity of calorimeter

First we need to calculate heat absorbed by calorimeter

Heat absorbed by calorimeter = heat lost by warm water – heat absorbed by cold water

                                              = 1966.48 J – 1799.12 J

                                              = 167.36 J

Change in temperature of calorimeter = 40.0 – 22.0

                                                     = 18 ^oC

Heat capacity of calorimeter = 167.36 J / 18 ^oC

                                             = 9.29 J/^oC