Question

4. The next question has to do with the condensation of water a. When water condenses on your glass, does it cool or warm the drink? b. If a can of soda at 1.0 °C has 6.512 grams of water condense on the outside, how much energy was transferred? The heat of condensation of water is 2.26kJ/g. c. If all the energy came from the soda, and the soda has the same heat capacity as water, what is the final temperature of the soda in °F?

5. A 5.69 g sample of copper metal was heated in boiling water to 99.8°C. Then it was dropped into a beaker containing 100.0 g of H2O at 22.6°C. Assuming the water gained all the heat lost by the copper, what is the final temperature of the H2O and Cu? The specific heat of water is CP=4.184 J/g∙°C and the specific heat of copper is CP,Cu=0.386 J/g∙°C

Answer 1

Solution of 4 (a)

It will warm drink by giving heat of condensation of water to drink.

Solution of 4 (b)

Heat Released from condensation of water = Mass of water (gram)* Heat of condensation(kJ/g)

                                                                        = 6.512*2.26 =14.717 kJ

Solution of 4 (C)

Heat capacity of water = 4.184 Joule/g °C

Assuming volume in soda can = 250 ml

                                                 = 250 g                       [ density = 1 gm/ml .. same as water]

Using Q = mC ΔT and assuming that no heat loss to surroundings. Hence, heat loss by water will be gain by soda Can.

                                                 ΔT= (14.717*1000) / (4.184*250) [ Q from part 4 (b)]

                                                 =14.069

                    Hence, Final temp = 1+ 14.069 = 15.069 °C

                                                 =59.124 °F

Solution of 5

After taking out from boiling water, temperature of Copper will be 99.8°C.

Let final temp of water and copper is T °C.

Hence, temp of copper will decrease from 99.8°C to T °C.

And temp of water will rise from 22.6°C to T °C.

Heat gained by water = heat lost by copper

Using Q = mC ΔT

100 (gm) * 4.184(Joule/g °C) * (T-22.6) = 5.69 (gm)*0.386 (Joule/g °C) *(99.8-T)

                                                 T = 23.0 °C.