More Specific Heat
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
What quantity of heat energy was lost?
Q = mcDT
= - 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
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 15°C to
Q = mcDT
= 200 000g (4.19)(60-15)
= 37 710 000J = 37 710 kJ
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)
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