For the following scenarios what is the metal? A piece of metal weighing 59.047 g was heated to 100.0 degree C and then put it into 100.0 mL of water (initially at 23.7 degree C). The metal and water were allowed to come to an equilibrium temperature, determined to be 27.8 degree C. Assuming no heat lost to the environment, calculate the specific heat of the metal. A 25.6 g piece of metal was taken from a beaker of boiling water at 100.0 degree C and placed directly into a calorimeter holding 100.0 mL of water at 25.0 degree C. The calorimeter heat capacity is 1.23 J/K. Given that the final temperature at thermal equilibrium is 26.2 degree C, determine the specific heat capacity of the metal.

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dfbustos dfbustos · Answer 1 · 2020-01-26T15:21:23+01:00

Answer:

a) [tex] C_{metal}=- \frac{100 gr * 4.186 \frac{J}{gr C} (27.8 -23.7)C}{ 59.047 gr *(27.8-100)C} =0.402 \frac{J}{ g C}[/tex]

b) [tex] C_{metal}= \frac{100 gr * 4.184 \frac{J}{gr C} (1.2)C +1.23 \frac{J}{K} (1.2 K))}{ 25.6 gr *(100-26.2)C} =0.267 \frac{J}{ g C}[/tex]

Explanation:

Part a

For this case we have the following data:

[tex] m_{metal}= 59.047 gr[/tex] the mass of the metal

[tex] c_{metal}=?[/tex] Is the value that we need to find

[tex] T_{f}= 27.8 C[/tex] represent the final temperature of equilibrium for the metal and the water

[tex] T_{i, metal}= 100 C[/tex] represent the initial temperature for the metal

[tex] m_{water}= 100 gr[/tex] since the density is 1g/ml

[tex] C= 4.186 \frac{J}{g C}[/tex] the specific heat for the liquid water

[tex] T_{i, water}= 23.7 C[/tex] the initial temperature for the water

For this case we have this equation:

[tex] \sum_{i=1}^n Q_i =0[/tex] if we have balance then we have this:

[tex] Q_{water} = -Q_{metal}[/tex]

And if we replace the formulas for heat we got:

[tex] 100 gr * 4.186 \frac{J}{gr C} (27.8 -23.7)C =- 59.047 gr * C_{metal}*(27.8-100)C[/tex]

And if we solve for [tex] C_{metal}[/tex] we got:

[tex] C_{metal}=- \frac{100 gr * 4.184 \frac{J}{gr C} (27.8 -23.7)C}{ 59.047 gr *(27.8-100)C} =0.402 \frac{J}{ g C}[/tex]

Part b

For this case we have the following balance:

[tex] Q_{lost} = Q_{gained}[/tex]

On this case the metal loss heat and this heat is gained by the water and the calorimeter, so we have this:

[tex] Q_{lost} = Q_{w} + Q_{calorimeter}[/tex]

And if we replace the info given we have this:

[tex] 25.6 gr *C_{metal} *(100-26.2)C = 100 gr * 4.184 \frac{J}{gr C} (26.2-25)C + 1.23 \frac{J}{K} (26.2+273 -25 -273)K[/tex]

And if we solve for [tex] C_{metal}[/tex] we got:

[tex] C_{metal}= \frac{100 gr * 4.184 \frac{J}{gr C} (1.2)C +1.23 \frac{J}{K} (1.2 K))}{ 25.6 gr *(100-26.2)C} =0.267 \frac{J}{ g C}[/tex]