Refer to Example 7-2 . The experiment is repeated with...

Refer to Example $7-2 .$ The experiment is repeated with several different metals substituting for the lead. The masses of metal and water and the initial temperatures of the metal and water are the same as in Figure $7-3$. The final temperatures are (a) $\mathrm{Zn}, 38.9^{\circ} \mathrm{C}$ (b) $\mathrm{Pt}, 28.8^{\circ} \mathrm{C} ;$ (c) $\mathrm{Al}, 52.7^{\circ} \mathrm{C}$. What is the specific heat capacity of each metal, expressed in $\mathrm{J} \mathrm{g}^{-1} \mathrm{C}^{-1} ?$

Refer to Example $7-2 .$ The experiment is repeated with several different metals substituting for the lead. The masses of metal and water and the initial temperatures of the metal and water are the same as in Figure $7-3$. The final temperatures are (a) $\mathrm{Zn}, 38.9^{\circ} \mathrm{C}$
(b) $\mathrm{Pt}, 28.8^{\circ} \mathrm{C} ;$ (c) $\mathrm{Al}, 52.7^{\circ} \mathrm{C}$. What is the specific heat capacity of each metal, expressed in $\mathrm{J} \mathrm{g}^{-1} \mathrm{C}^{-1} ?$

Key Concepts

Specific Heat Capacity

Specific heat capacity is a measure of the amount of heat energy required to raise the temperature of a unit mass of a substance by one degree Celsius (or one Kelvin). It is an intrinsic property of a material that helps in understanding how efficiently that material stores and releases heat energy.

Calorimetry

Calorimetry is the experimental method used to measure the amount of heat transferred in a chemical or physical process. By carefully controlling the conditions in a calorimeter, one can determine the heat absorbed or released by substances as they undergo temperature changes, which is essential for calculating properties like specific heat capacity.

Conservation of Energy in Thermal Processes

The principle of conservation of energy is central to thermal experiments, especially those involving heat exchange. In a closed system, the heat lost by a hotter substance must equal the heat gained by a cooler one when the system reaches thermal equilibrium. This balance allows for the determination of unknown thermal properties by equating the heat exchange between substances.

Transcript

00:01 Okay, so we want to calculate the specific heat for a given number of metals.

00:08 And if you go by figure 7 .3 in the textbook, you can see what exactly the experiment is all about.

00:20 I will also explain it over here very briefly.

00:24 So you have first case or first scenario where you have a test tube.

00:33 Immersed in water and this is on some kind of a heat heat bath or water bath or whatever okay so you have 150 grams of metal inside this test tube and you're heating it this is scenario number one and then you have the second case where you have an insulated situation this is a thermometer and you have water okay and this is 15 grams of water okay and this is all insulated and the temperature reads here is 22 degrees celsius so this is the temperature that we have for water and then you have a third scenario where you again have an insulated system system okay and in this case you have the metal metal inside an insulated system immersed in water and there is some final temperature which is tf okay so what you need to find is you have to find the specific heat capacity okay for the first one is zinc and they have given the initial temperatures and they have given the final temperatures so in this case the final temperature is 38 .9 degrees of celsius you have platinum in which case the final temperature is 20 is 28 .8 degrees celsius and then you have third case where you have aluminium final temperature is 52 .7 degrees celsius okay so we'll start from the first case so in this case you have a metal which is zinc and the final temperature is equal to 38 .9 degrees celsius so we have to do exactly the same thing one thing that you need to remember is the heat gained by water it's equal to the heat that is lost by the metal okay so we'll calculate the heat um gained by water which is q h2 o is equal to you first have to find the mass so in this case you have 50 grams of water right let's just see okay so this is what we'll find here we gain by water so the formula to do this is m cp delta t so m is the mass which is equal to 50 grams cp is the specific heat capacity and this is 4 .18 jules per gram per degree celsius and then delta t so in this case the initial temperature was 22 .0 and the final temperature is 38 point 9.

04:47 Okay.

04:48 So you have final temperature as 38 .9.

04:52 Initial temperature is 22 degrees.

04:55 Okay.

04:56 So once you do this, you get 3 .5 into 10 to the power 3 joules.

05:03 Now from this equation number 1, so from equation 1, q zinc is equal to minus q of water.

05:26 So now this value will be minus 3 .5 into 10 to the power 3 joules.

05:33 So now i substituted...