Answers to p118

More Specific Heat

Basic Problems

1. In the summertime, you find that tap water at 18C is too warm to drink. You put

500 mL of this water in the refrigerator. After a period of time, the temperature of

the water is 4C. While it was cooling, the water lost a certain quantity of heat


What quantity of heat energy was lost?

Q = mcDT

= 500(4.19)(4-18)

= - 29330 J = 29.33 kJ

2. A calorimeter contained 250 g of water at 24C. An electric current was passed

through a heater placed in the water. The heater transferred 14 700 J of energy to

the water.

What is the final temperature of the water?

Q = mcDT

14 700 J = 250 (4.19)(x-24)

x = 38.0 oC

3. A water tank contains 200 kg of water. The water is heated by a 4500-W heating


How much energy is required to raise the temperature of the water from 15C to


Q = mcDT

= 200 000g (4.19)(60-15)

= 37 710 000J = 37 710 kJ

Mixing Problems

4. What mass of copper, originally at 50.0 oC, must be added to 1.0 kg of 10.0 C

water to raise its temperature to 20.0 oC? [ sp heat for Cu = 0.39 J/(g oC) ]

-Qlost by copper = Q gained by water

- mcDT for copper= mcDT for water

-m(0.39)(20.0 50)= 1000g(4.19)(20.0 - 10)

m = 3581 g

5. A 450 mL sample of water is originally at 25.0 C. How cold will it get if we add

300 mL of 0.5 oC water to that sample?

-Qlost by warm water = Q gained by cold water

- mcDT for warm water = mcDT for cold water

-450(4.19)(x - 25) = 300(4.19)(x 0.5)

x = 15.2 oC

6. Zinc(Zn), platinum(Pt) and titanium(Ti) follow the Mc = 25 formula for metals,

where M is the molar mass in g/mole and c = specific heat in J/(goC). Estimate the

specific heat for these three elements.

To get c = 25/M, simply look up the molar mass of each metal. For example, Zn = 65.39 g/mole

c = 25/65.39 = 0.39 J/[g C];

Pt = 0.13 J/[g C]; Ti = 0.52 J/[g C];

7. Based on the relationship Mc = 25 J/(mole oC), what elemental metal has the

highest specific heat? M = molar mass in g/mole

Since c = 25/M, the metal with the lowest molar mass will have the highest specific heat. That metal is lithium.


8. Starting with the same formula used in #4, prove that for equal masses of the

same material, the final temperature obtained by mixing two samples will simply

be the average of the two initial temperatures.

-Qlost by hot material = Q gained by cold material

- mcDT for hot material = mcDT for cold material

but if m and c are equal, they cancel and we are left with

-(Tf Ti hot) = (Tf Ti cold)

-Tf + Ti hot = Tf Ti cold

Ti hot + Ti cold = Tf + Tf

Ti hot + Ti cold = 2 Tf

(Ti hot + Ti cold)/2 = Tf