How to Determine the Specific Heat of a Substance
Problems #1 - 10

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Problem #1: Suppose a piece of iron with a mass of 21.5 g at a temp of 100.0 °C is dropped into an insulated container of water. The mass of the water is 132.0 g and its temperature before adding the iron is 20.0 °C. What will be the final temp of the system? Specific heat of iron is 0.449 kJ/kg K.

Solution:

1) Since

qlost, metal = qgained, water

we write

(mass) (Δt) (Cp, metal) = (mass) (Δt) (Cp, water)

2) Substituting:

(21.5) (100 - x) (0.449) = (132.0) (x - 20) (4.184)

Some explanation:

a) 100 - x is the Δt for the metal; it starts at 100.0 °C and drops to some unknown, final value.
b) x - 20 is the Δt for the water; it starts at 20.0 °C and rises to some unknown, final value.
c) Since both metal and water wind up at the same ending value, we need to use only one unknown for the two Δt expressions.

3) A wee bit of algebra:

(2150 - 21.5x) (0.449) = (132x - 2640) (4.184)

965.35 - 9.6535x = 552.288x - 11045.76

561.9415x = 12011.11

To 3 sig figs, the answer is 21.4 °C.


Problem #2: A 12.48 g sample of an unknown metal, heated to 99.0 °C was then plunged into 50.0 mL of 25.0 °C water. The temperature of the water rose to 28.1 °C. Assuming no loss of energy to the surroundings:

1. How many joules of energy did the water absorb?
2. How many joules of energy did the metal lose?
3. What is the heat capacity of the metal?
4. What is the specific heat of metal?

Solution:

1) q = (50.0 g) (3.1 °C) (4.184 J g-1 °C-1) = 648.52 J

2) 648.52 J

3) 648.52 J / 70.9 °C = 9.147 J/°C

4) 9.147 J/°C divided by 12.48 g = 0.733 J g-1 °C-1

Comment #1: this question doesn't use the qlost = qgained formulation of other questions. That is because the question is broken up into four parts. Notice that parts (1) and (2) are the equivalent of qlost = qgained and that (4) is the usual answer sought in problems of this type.

Comment #2: (3) is a step unnecessary to the solution for (4). It is there so you notice the difference between heat capacity and specific heat capacity.


Problem #3: A 43.2 g block of an unknown metal at 89.0 °C was dropped into an insulated vesssel containing 43.00 g of ice and 26.00 g of water at 0 °C. After the system had reached equilibrium it was determined that 9.15 g of the ice had melted. What is the specific heat of the metal? (The heat of fusion of ice = 334.166 J g¯1.)

Solution:

Comment: this variation of the usual suspects (detailed above) does NOT involve a temperature change in the water, only in the metal. Rather, some ice melts and the whole ice-water system stays at zero Celsius. Verrrry interesting!

1) Determine heat gained by the ice that melted:

9.15 g times 334.166 J g¯1 = 3057.62 J

2) Substitue and solve for the specific heat:

q = (mass) (Δt) (Cp, metal)

3057.62 J = (43.2 g) (89.0 °C) (x)

x = 0.795 J g¯1 °C¯1


Problem #4: A 35.0 g block of metal at 80.0 °C is added to a mixture of 100.0 g of water and 15.0 g of ice in an isolated container. All the ice melted and the temperature in the container rose to 10.0 °C. What is the specific heat of the metal?

Solution:

1) Determine heat required to melt the ice:

q = (15.0 g) (334.166 J g¯1) = 5012.49 J

Note that the 100 g of water is not mentioned yet.

2) Determine heat need to raise 115 g of water from 0 to 10.0 °C:

q = (115 g) (10.0 °C) (4.184 J g¯1 °C¯1) = 4811.6 J

Note the inclusion of the melted 15 g of ice. Also, notice that the water was at zero °C. We know this from the presence of the ice.

3) Determine the specific heat of the metal:

(5012.49 J + 4811.6 J) = (35.0 g) (70.0 °C) (x)

x = 4.01 J g¯1 °C¯1


Problem #5: A 500.0 g sample of an element at 153.0 °C is dropped into an ice-water mixture. 109.5 g of ice melts and there is still an ice - water mixture. What is the specific heat of the metal in J/g-°C? Given the molar heat capacity of the metal is 26.31 J/mol °C, what is the atomic weight and identity of the metal?

Solution:

1) Determine energy needed to melt the ice:

(6.02 kJ/mol) (109.5 g / 18.015 g/mol) = 36.5912 kJ

2) Determine specific heat:

36591.2 J = (500.0 g) (153.0 °C) (x)

x = 0.4783 J/g-°C

Note: we know the change in temperature is 153.0 °C because there is still ice in the water. That means the ice-water mix remained at zero Celsius as the 109.5 g of ice melted.

3) Determine aomic weight of the element:

0.4783 J/g-°C times x = 26.31 J/mol °C

x = 55.0 g/mol

The element is manganese.


Problem #6: A 12.48 g sample of an unknown metal is heated to 99.0 °C and then was plunged into 50.0 mL of 25.0 °C water. The temperature of the water rose to 28.1°C.

(a) How many joules of energy did the water absorb?
(b) How many joules of energy did the metal lose?
(c) What is the heat capacity of the metal?
(d) What is the specific heat capacity of the metal?

The definitions for heat capacity and specific heat capacity may be found here.

1) Solution to (a):

q = (50.0 g) (3.1 °C) (4.181 J g¯1 °C¯1) = 648.52 J

I used 50.0 g because the density of water is 1.00 g/mL and I had 50.0 mL of water.

2) Solution to (b):

q = 648.52 J

We assume all heat absorbed by the water was lost by the metal. We assume no loss of heat energy to the outside during the transfer.

3) Solution to (c):

648.52 J / 74.0 °C = 8.76 J/°C (or 8.76 J/K)

4) Solution to (d):

(50.0 g) (3.1 °C) (4.181 J g¯1 °C¯1) = (12.48 g) (74.0 °C) (x)

Solve for x.


Problem #7: What is the specific heat of a metal if addition of 90.0 g of the metal at 17.7 °C to 210.0 g of Cu (s = 0.385 J/g-°C) at 153.7 °C produces a mixture that reaches thermal equilibrium at 129.1 °C?

Solution:

Comment: notice that the two metals are being added to each other. Imagine a situation where each sample is composed of dust or very small pellets. Then, the two dry samples are rapidly mixed together.

(90.0 g) (111.4 °C) (x) = (210.0 g) (24.6 °C) (0.385 J/g-°C)

x = 0.198 J/g-°C


Problem #8: A 31.0 gram block of an unknown metal at 88.0 °C was dropped into an insulated flask containing approximately 30.0 grams of ice and 20.0 grams of water at 0.0 °C. After the system had reached a steady temperature, it was determined that 12.1 grams of ice had melted. What is the specific heat of the metal? The heat of fusion of ice is equal to 334.166 J/g.

Solution:

12.0 g times 334.166 J/g = 4009.992 J

4009.992 J = (31.0 g) (88.0 °C) (x)

x = 1.47 J / g °C

Comment: the fact that ice remained in the water when the temperature reached equilibrium means the water-ice mixture stayed at zero Celsius. This means the metal went from 88.0 °C to 0 °C, for a Δt of 88.0 °C


Problem #9: A 25.95 g sample of methanol at 35.60 °C is added to a 38.65 g sample of ethanol at 24.70 °C in a constant-pressure calorimeter. If the final temperature of the combined liquids is 28.65 °C and the heat capacity of the calorimeter is 19.3 J/C, determine the specific heat of methanol.

Solution:

the heat lost by the methanol goes to (1) heating the ethanol and (2) heating the calorimeter

(25.95 g) (6.95 °C) (x) = (38.65 g) (3.95 °C) (2.44 J g-1 °C-1) + (3.95 °C) (19.3 J/C)

x = 2.49 J g-1 °C-1


Problem #10: A student heats a piece of 130 g of an unkown grayish metal to a temp. of 99.2 °C. she places the metal in a styrofoam cup that contains 55.7 g water at a temperature of 23.0 °C. The hot metal heats the water in the cup to 31.4 °C.

a) Calculate the specific heat of the metal.
b) What is the atomic weight?
c) Identify the metal.

Solution:

q = (55.7 g) (8.4 °C) (4.184 J g-1 °C-1 = 1957.61 J

1957.61 J = (130 g) (67.8 °C) (x)

sp ht. = 0.222 J g-1 °C-1

Use the Dulong–Petit law:

M = 3R / sp. ht

M = (3) (8.31446 J mol-1 K-1) / 0.222 J g-1 K-1

M = 112 g /mol

Cadmium

Note the shift from °C to K. This is allowed because the "size" of one °C is equal to the "size" of one K.


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