**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 ^{o}C **

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) ]

-**Q _{lost}**

**- 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?

-**Q _{lost}**

**- mcDT for **_{warm water }= mcDT for _{ }cold water

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

6. Zinc(Zn), platinum(Pt) and titanium(Ti) follow the M*c *= 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 M*c *= 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.

-**Q _{lost}**

**- mcDT for **_{hot material }= mcDT for **cold material**

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

-(T_{f} – Ti _{hot}) = (T_{f} – Ti _{cold})

-T_{f} + Ti _{hot } = T_{f} – Ti _{cold}

Ti _{hot } + Ti _{cold} = T_{f}_{ }+ T_{f}

Ti _{hot } + Ti _{cold} = 2 T_{f}_{ }

(Ti _{hot } + Ti _{cold})/2 = T_{f}_{ }