Solve Temperature/Metal Help: Calc Specific Heat & ID

[drscrewyou]

[SOLVED] Temperature/Metal Help

m in desperate need for help because I am horrible at chemistry and i need to do some of these prelab questions that i do not understand.

1. How do you know the final temperature of the metal?

2. Why do you use a hot water bath to heat the metal? (obvious but i think there's something more in depth)

3. Why do you measure the volume of the water in the calorimeter when we need to know the mass of the water for the calculations.

4. *Kinda Long* - You ar given a metal and asked to determine its identity. You are to do this by determining the specific heat of the metal. You place the metal in a boiling water bath for a few minutes and then transfer the metal to a 100.0 g sample of water at a measured temperature. You then record the highest temperature of the water. The dta are given in the table below.

Mass water - 100.0 g
Initial temp of water 21.31 degrees C
FInal temperature of water 24.80 degrees C
Mass metal 50.0 g


a. Given that the specific heat capacity of water is 4.184 J/g C, calculate the amount of heat that is absorbed by the water.

b. how does the heat absorbed by the water compare to the heat lost by the metal?

c. What is the initial temperature of the "hot" metal? What is the final temperature of the metal?

d. Calculate the specific heat capacity of the metal.

e. Given the following specific heat capacities, determine the identity of your metal.

Tin: 0.227 J/g C
Zinc: 0.388 J/g C
Aluminum: 0.891 J/g C

Thank You Fellow Chemistry Members

 

[Telmerk]

If all the heat from the metal can only pass into the water:

1. c_metal *_m_metal *delta_T_metal = c_water * m_water * delta_T_water

2. delta_T_water = T_final_water - T_initial_water and so on in the case of the metal

From this I think it is possible to answer the 4. question.

Regarding the final temp. of the metal (I am not so good in English, so maybe I misunderstood sg ) I think it should be equal to the final temp. of water when reaching equilibrium (i.e. no more heat transfer).

Cheers, TtM