More Specific Heat

Basic Problems

1. In the summertime, you find that tap water at 18°C 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 4°C. While it was cooling, the water lost a certain quantity of heat

energy.

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 24°C. 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

element.

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

60°C?

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.

Challenger

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