Question

At a local convenience store, you purchase a cup of coffee, but, at 98.4°C, it is too hot to drink. You add 34.3 g of ice that is −2.2°C to the 248 mL of coffee. What is the final temperature of the coffee? (Assume the heat capacity and density of the coffee are the same as water and the coffee cup is well insulated.)

Answer 1

Assuming that the coffee cup is insulated,

heat lost by hot coffee=heat gained by ice

Density of pure water=1 g/mL=density of coffee (as given in the question)

Mass of coffee=Density x Volume=1 g/mL x 248 mL=248 g

Heat capacity of pure water=4.184 J/g°C=heat capacity of coffee (as given in the question)

Heat capacity of ice=2.09 J/g°C

Heat of fusion of ice=333.55 J/g

Let the final temperature attained by the system be T_f

Initial temperature of coffee=98.4°C

Initial temperature of ice=-2.2°C

Heat lost by coffee=mass of coffee x specific heat of coffee x fall in temperature of coffee

=248 g x 4.184 J/g°C x (98.4°C - T_f)

=102102.99 J - 1037.63 J/°C x T_f

First the temperature of ice increases from -2.2°C to 0°C then it melts at 0°C and then its temperature increases to T_f

Heat gained by ice=mass of ice x specific heat of ice x rise in temperature of ice+mass of ice x heat of fusion of ice + mass of ice x specific heat of the water x increase in temperature of water

=34.3 g x 2.09 J/g°C x (0°C - (-2.2°C))+34.3 g x 333.55 J/g + 34.3 g x 4.184 J/g°C x (T_f​​​​​​ - 0°C)

=34.3 g x 2.09 J/g°C x 2.2°C + 34.3 g x 333.55 J/g + 34.3 g x 4.184 J/g°C x (T_f -0°C)

=157.71 J + 11440.77 J + 143.51 (T_f)

=11598.48 J + 143.51 J/°Cx T_f

Heat lost by hot coffee = heat gained by ice

102102.99 J -1037.63 J/°C x T_f=11598.48 J +143.51 J/°C x T_f

102102.99 J - 11598.48 J=143.51 J/°C x T_f +1037.63 J/°C x T_f

90504.51 J=1181.14 J/°C x T_f

T_f=90504.51 J/1181.14 J/°C=76.62°C

So final temperature of the coffee=T_f=76.62°C