• Qlost Qgained
• Calorimetry Simulation
• Calorimeter Simulation
• Calorimetry Simulation Lab
• Measuring Specific Heat

Q gained: 145.04J, Q lost: 94.51J, Q retained net: 50.53J ∆T (max) = 54.8-49.9 = 4.9°C Below is the completed project, like an answer key, with sample data and calculations in. Recall that Q is the amount of heat lost or gained. We want to know the final temperature of the system, and we have determined that occurs when the amount of heat transferred from the water to the bucket equals the amount of heat that is transferred from the bucket to the water. Strictly speaking, a positive value for Q should indicate a gain of heat (since T 2 - T 1 is a positive number when there is an increase in temperature) or a greater amount of the looser phase, and a negative value for Q when there is a temperature loss or greater amount of the more tightly - bound phase.

by

Marilynn Stone

This SMART website is hosted by the Illinois Institute of Technology

Most text books are very confusing when it comes to solving heat problems. I have discovered an easier way to do these problems.

Just use: Heat lost + Heat gained = 0

If you use this simple equation, ΔT is always T_f - T_i, and you don't have to figure out which substances are losing heat and which are gaining heat, just plug in all the numbers and solve.

Examples:

1. A metal sample with a mass of 0.0500 kg at 100⁰C is placed into 0.400 kg a water at 20.0⁰C. If the final temperature is 22.1⁰C, what is the specific heat of the metal sample? (Assume that there is no heat lost to the surroundings.)

m_m = 0.500 kg

T_im = 100⁰C

T_f = 22.1⁰C

m_w = 0.400 kg

T_iw = 20⁰C

T_fw = 22.1⁰C

c_w = 4186 J/kg·⁰C

c_m = ?

Heat lost + Heat gained = 0

Q = mcΔT

m_mc_mΔT_m + m_wc_wΔT_w = 0

Plug in the numbers:

0.0500 kg(c_m)(22.1⁰C - 100⁰C) + 0.400 kg(4186 J/kg·⁰C⁾⁽22.1⁰C - 20⁰C) = 0

Calculate the change in temperautres:

0.0500 kg(c_m)(-77.9⁰C) + 0.400 kg(4186 J/kg·⁰C)(2.1⁰C) = 0

Multiply:

-3.895 kg·⁰C(c) + 3516.24 J = 0

Add 3.895 kg·⁰C(c) to both sides:

3516.24 J = 3.895 kg·⁰C(c)

Divide both sides by 3.895 kg·⁰C:

c = 902.76 J/kg·⁰C

2. If 0.022 kg of ice at 0⁰C is added to 0.450 kg of water at 80⁰C, what is the final temperature. (Assume that there is no heat lost to the surroundings.)

m_I = 0.022 kg

T_iI = 0⁰C

L_f = 3.33 x 10⁵ J/kg

m_w = 0.450 kg

T_iw = 80⁰C

c_w = 4186 J/kg·⁰C

T_f = ?

Heat lost + heat gained = 0

Q = mL_f Q = mcΔT

The ice has to melt first before the final temperature can be found.

m_IL_f + m_Ic_wΔT_I + m_wc_wΔT_w = 0

Plug in the numbers:

0.022 kg(3.33 x 10⁵ J/kg) + 0.022 kg(4186 J/kg·⁰C)(T_f - 0⁰C) + 0.450 kg(4186 J/kg·⁰C)(T_f - 80⁰C) = 0

Multiply:

7326 J + 92.092 J/⁰C(T_f) - 0⁰C + 1883.70 J/⁰C(T_f) - 150696 J = 0

Qlost Qgained

Combine like terms:

1975.792 J/⁰C(T_f) -143370 J = 0

Add 143370 J to both sides:

1975.792 J/⁰C(T_f) = 143370 J

Divide both sides by 1975.792 J/⁰C

T_f = 72.56⁰C

3. A copper sample with a mass of 0.312 kg is at 99⁰C. The sample is placed into 0.507 kg of water at 21⁰C which is in an aluminum calorimeter which has a mass of 0.105 kg. If the final temperature is 25⁰C, calculate the specific heat of copper. (Assume that there is no heat lost to the surroundings.)

m_c = 0.312 kg

T_ic = 99⁰C

T_fc = 25⁰C

c_c = ?

m_w = 0.507 kg

T_iw = 21⁰C

T_fw = 25⁰C

c_w = 4186 J/kg·⁰C

m_a = 0.105 kg

T_ia = 21⁰C

T_fa = 25⁰C

c_a = 889 J/kg·⁰C

Heat lost + Heat gained = 0

m_cc_cΔT_c + m_wc_wΔT_w + m_ac_aΔT_a = 0

Plug in the numbers:

0.312 kg(c_c)(25⁰C - 99⁰C) + 0.507 kg(4186 J/kg·⁰C)(25⁰C - 21⁰C) + 0.105 kg(889 J/kg·⁰C)(25⁰C - 21⁰C) = 0

Calculate the change in temperatures:

0.312 kg(c_c)(-74⁰C) + 0.507 kg(4186 J/kg·⁰C)(4.0⁰C) + 0.105 kg(889 J/kg·⁰C)(4.0⁰C) = 0

Multiply:

-23.088 kg·⁰C(c_c) + 8489.208 J + 373.38 J = 0

Combine like terms:

-23.088 kg·⁰C(c_c) + 8862.588 J = 0

Add 23.088 kg·⁰C(c_c) to both sides:

8862.588 J = 23.088 kg·⁰C(c_c)

Divide both sides by 23.088 kg·⁰C:

c_c = 384 J/kg·⁰C

( the accepted value for copper is 387 J/kg·⁰C)

4. To what temperature must a 0.853 kg block of lead be heated in order to completely melt 0.015 kg of ice which is initially at -5⁰C. The final temperature of the melted ice is +5⁰C. (Assume that there is no heat lost to the surroundings.)

m_L = 0.853 kg

c_L = 128 J/kg·⁰C

T_iL = ?

T_f = 5⁰C

m_I = 0.015 kg

c_I = 209 J/kg·⁰C

T_iI = -5⁰C

T_fI = 0⁰C

L_f = 3.33 x 10⁵ J/kg

c_w = 4186 J/kg·⁰C

T_iw = 0⁰C

T_fw = 5⁰C

Heat lost + Heat gained = 0

First the ice must heat up to 0⁰C Q = m_Ic_IΔT_I

Then the ice has to melt Q = m_ILf

Next the melted ice has to warm up to 5⁰C Q = m_Ic_wΔT_w

m_Lc_LΔT_L + m_Ic_IΔT_I + m_ILf + m_Ic_wΔT_w = 0

Plug in the numbers:

0.853 kg(128 J/kg·⁰C)(5⁰C - T_i) + 0.015kg(209 J/kg·⁰C )[0⁰C - (-5⁰C)] + 0.015kg(3.33 x 10⁵J/kg) + 0.015 kg(4186J/kg·⁰C )(5⁰C - 0⁰C) = 0

Multiply:

545.92 J - 109.184 J/⁰C(T_i) + 0 J + 15.675 J + 4995 J + 313.95 J - 0 J = 0

Combine like terms:

5870.545J -109.184 J/⁰C(T_i) = 0

Add 109.184 J/⁰C(T_i) to both sides:

Divide both sides by 109.184 J/⁰C

T_i = 53.8⁰C

Back to the SMART home page.

Calorimetry Simulation

Calorimetry is the measurement of the amount of heat gained or lost during some particular physical or chemical change. Heats of fusion or vaporization, heats of solution, and heats of reaction are examples of the kinds of determination that can be made in calorimetry. The term itself derives the Latin word for heat, caloric, as is the name of the instrument used to make these determinations, the calorimeter.

Calorimetry is the science associated with determining the changes in energy of a system by measuring the heat exchanged with the surroundings. A calorimeter is a device used to measure the quantity of heat transferred to or from an object. In calorimetry it is often desirable to know the heat capacity of the calorimeter itself rather than the heat capacity of the entire calorimeter system (calorimeter and water). The heat (Q) released by a reaction or process is absorbed by the calorimeter and any substances in the calorimeter.

Results & Discussions~

Table 6-A Determination of the amount of heat energy transferred using hot and cold water.

Calorimeter Simulation

For this part of the experiment, we are comparing the heat gained and the heat lost by the system. Since it is in the calorimeter, we assume that the system is isolated. Which means that the heat gained is equal to the heat lost ( Q _gained = Q _lost ). Using the specific heat of water, C = (1 cal/g-C^o) we compute the heat gained and the heat lost by using the equations: Q _{gained by the cold water} = m _{cold water} C (T _final – T _{cold water} ) and Q _{lost by the hot water} = m _{hot water} C (T _{hot water} – T _final ). All the necessary data are gained by measuring from the actual experiment. After some computation and doing these with three trials, we got a percent differences of 7.43 % , 0.71 % , and 4.26%.

Table 6 – B : Determination the Specific Heat of Materials.

On the second part, we are getting the specific heat of the aluminum, copper, and lead. To do so, we are given a sample of the metals and we are getting the data from the steps that was stated in the manual, and when the necessary data are gained, we can susbstitute it to the equation : m _metal ( C _{p metal} ) ( T _metal – T _final ) = m _water ( C _{p water} ) ( T _final – T _{cold water} ). We still use C _pwater = 1 cal/g-C ^o . As a result, we got 47.9 % , 92.25 % , and 58.67 % from the metals that we compared to the theoretical value of the specific heats. These are relatively high because of some factors like, the room temperature, measurement-error, human error , and other factors.

Summary~

To sum up, the amount of heat, Q, required raising the temperature of a solid body or material at constant pressure depends on the change in temperature ∆T of the material, its mass (m), and a characteristic of the material forming the body called its specific heat, c. This relationship is expressed by the equation Q = mc∆T and the dimensions of C are thus heat per unit mass per unit temperature change. We know that when two bodies, initially at different temperatures, are placed in intimate contact, in time they will come to equilibrium at some intermediate temperature. Provided no heat is lost to or gained from the surroundings, the quantity of heat lost by the hotter body is equal to that gained by the colder body. This is the process which occurs in the method of mixtures that you will used (Bauer, 2014).

In our experiment, adding hot water to cold water wasn’t the same result as add cold water to hot water because of the heat is transferred from high temperature to low temperature. Heating up a materials caused expanding, speeding up and pushing further apart of its molecules. Also hot water has more thermal energy before we were mixed them than after we mixed them and they conserved the energy because of the heat lost and gained are almost equal.

On the other hand, the specific heat capacity of any metals is lower compared to the specific heat capacity of water.

Conclusion~

First of all, calorimeter was not completely isolated and heat was lost also before the hot water was poured into calorimeter and the room temperature (air-conditioned) is the one which affects the values of our results. Meanwhile, in the experiment determining the specific heat of metals, we got higher percentage errors these are due to heat loss to the surroundings while we are trying to transfer the hot metal objects, boiling point of metals are too long, room temperature, and apparatuses. However we can conclude that the specific heat of metals is lower than the specific heat of water.

Problem Solving~

Number 1

Calorimetry Simulation Lab

Number 2

Measuring Specific Heat

Number 3