Chemistry Exam 2

An unknown metal absorbs 61 J of heat, and its temperature increases by 29∘C. What is the heat capacity of the metal?

Recall that heat capacity, C, is the ratio of heat to the change in temperature. C=qΔT C=(61J)(29∘C) =2.103J∘C The answer should have two significant figures, so round to 2.1J∘C

The transfer of thermal energy between two objects at different temperatures is called which of the following?

heat

The equation that contains spectator ions is called the:

The complete ionic equation- contains all the ions in solution, including spectator ions.

Joule per gram-degree Celsius is a unit of which of the following?

Specific heat: The specific heat capacity (c) of a substance, commonly called its "specific heat," is the quantity of heat (in Joules) required to raise the temperature of 1 gram of a substance by 1 degree Celsius.

An object absorbs 1.70×104J of heat and as a result, its temperature is increases from 45.0∘C to 75.0∘C. What is its heat capacity of the object?

Recall that heat capacity, C, is the ratio of heat to the change in temperature. C=qΔT First calculate ΔT using the following formula: ΔT=Tfinal−Tinitial ΔT=75.0∘C−45.0∘C=30.0∘C C=1.70×104 J/30.0∘C=566.67 J∘C Finally, after rounding the answer to three significant figures, we find that the heat capacity of the object is 567J∘C

What will be the final temperature of a 207.0 g piece of copper (specific heat = 0.385Jg∘C) that absorbs 5.00 kJ of heat starting at 80.4∘C? Report your answer with the correct number of significant figures.

The final temperature can be determined by rearranging the equation for specific heat capacity for ΔT and substituting the known values of mass, specific heat, and heat absorbed. qcm=ΔT (5,000. J)(0.385Jg∘C)(207.0 g)=62.7∘C The positive sign of ΔT makes sense, as we know that energy is absorbed in this endothermic process: (Tinitial+ΔT=Tfinal) 80.4∘C+62.7∘C =143.1∘C

How many grams of water can be heated from 5.0∘C to 37.0∘C with 20.0J of heat? The specific heat of water is 4.184Jg∘C. Round to three significant figures.

The mass can be calculated using the following equation: m=qc×ΔT. m=20J4.184Jg∘C×(37.0∘C−5.0∘C) =20J4.184Jg∘C×(32.0∘C) =0.149g (three significant figures).

Which amount of water would have the greatest heat capacity?

The ocean: The ocean is an extremely large body of water, so it would take an immense amount of heat energy to be dispersed around all of the water molecules in order to raise the temperature of the entire body of water by one degree Celsius.

Calculate the heat capacity of 18.5 g iron given that 2,982 J is needed to raise the temperature by 194 ∘C.

We can calculate the heat capacity of the 18.5g sample of iron by dividing the heat by the change in temperature. C=qΔT =2,982 J/194 ∘C =15.4J∘C Therefore, after rounding the answer to three significant figures, the heat capacity is about 15.4J∘C.

The specific heat of water is 4.184Jg ∘C. Determine the final temperature when 600.0 g water at 75.5∘C absorbs 5.90×104 J of energy. Report your answer with three significant figures.

Use the equation q=mcΔT to solve for the change in temperature. q=mcΔT ΔT=q/mc ΔT=(5.90×104J/4.184Jg∘C)⋅600.0g ΔT=23.5∘C Now, recognize that ΔT=Tf−Ti to solve for the final temperature. ΔT=Tf−Ti Tf=ΔT+T Tf=23.5∘C+75.5∘C Tf=99.0∘C

How much heat, in kilojoules, must be added to a 580 g aluminum pan to raise its temperature from 25∘C to 150∘C? The specific heat capacity for aluminum is 0.897Jg∘C. Round the answer to two significant figures.

Use the equation shown below, where c is the specific heat of aluminum, 0.897, m is the mass of the pan, 580 g, the initial temperature is 25∘C, and the final temperature is 150∘C. q=(580 g)(0.897Jg∘C)(150∘C−25∘C) q=65,032.5 J Converting to kilojoules, and rounding to two significant figures, q=65 kJ.

Heat capacity depends on:

Heat capacity is an extensive property in that it depends on the composition and amount of a substance. For example, a metal wire tends to have a lower heat capacity than most amounts of water; a wire heats up and cools down much faster than water (i.e., it depends on the substance's composition). Also, a small amount of water has a lower heat capacity than a large amount of water. The ocean has a relatively stable amount of heat, while heat changes dramatically in a puddle (i.e., it depends on the amount of substance). This is different from the specific heat capacity, which is an intensive property and is independent on the amount of substance and depends only on the composition of the substance.

If an object has a specific heat of 0.981 Jg∘C, absorbs 674 J of heat, and experiences a 12.0∘C temperature increase, what is the object's mass in grams?

The specific heat capacity (c) is the quantity of heat required to raise 1 gram of the material 1 degree Celsius. c=qm/ΔT Solve the equation for m to find the mass. m=qc/ΔT =674J/0.981 Jg∘C×12.0∘C =57.2545g Since the answer should have three significant figures, round to 57.3g

A 1.0 kg object absorbs 1,303 J of heat energy and experiences a temperature increase of 5.2∘C. What is the object's specific heat, in joules per gram-degree celsius? Report your answer with the correct number of significant figures.

The specific heat capacity (c) is the quantity of heat required to raise 1 gram of the material 1 degree Celsius: c=qmΔT First, convert the mass to grams. 1.0kg×1000.g1.0kg=1000.g Now use the equation to find c. c=(1,303J*1000.g)/5.2∘C =0.2506Jg∘C Since the answer should have two significant figures, round to 0.25Jg∘C

If 202 J is absorbed by a 51 g sample and the temperature increases from 25.2∘C to 29.1∘C, what is the specific heat? Report your answer with two significant figures.

c=qmΔT c=202 J51 g(3.9∘C) c=1.0Jg∘C

A calorie is equal to the amount of energy required to:

raise the temperature of one gram of water by one degree Celsius (The calorie is based on one gram of water and the Celsius temperature scale.)