How Is Thermal Energy Absorbed in a Calorimeter Experiment with Tin?

[e.hill792]

Homework Statement

Find the quantity of thermal energy absorbed but the water in the calorimeter.
Mass of metal sample:69.64g
Water sample: Volume 190mL
Temp: 24'C
boiling water: 275mL
temp of boiling water:90'C
Sample in Calorimeter
Initial Heat: 25'C
30 seconds: 26'C
60 seconds 26'C
90 seconds 26'C
120 seconds 26'C
2nd Trial
Mass of metal sample:69.64g
Water sample: Volume 170mL
Temp: 26'C
boiling water: 275mL
temp of boiling water:98'C
Sample in Calorimeter
Initial Heat: 31'C
30 seconds: 29'C
60 seconds 29'C
90 seconds 28'C
120 seconds 28'C

Homework Equations

(m_H2O)(c_H2O)(deltaT_H2O)=-(m_metal)(c_metal)(deltaT_metal)

The Attempt at a Solution

(190)(4.18)(-66)=(69.64)(c_metal)(-3)
______________ =250.89 c_metal
(69.64)(-3)

(170)(4.18)(-69)=(69.64)(c_metal)(1)
______________= 704.06
(69.64)(1)

I know the sample is tin because the metal was labeled "Sn" even though the point of the lab was to "identify" an unknown metal.

 

[nate808]

Hello,

Thank you for sharing your data and attempt at a solution. Based on the given information, I can calculate the quantity of thermal energy absorbed by the water in the calorimeter.

First, we need to calculate the change in temperature for the water sample in each trial. In the first trial, the initial temperature of the water sample was 24°C and after 120 seconds, it reached 26°C. Therefore, the change in temperature is 2°C. In the second trial, the initial temperature of the water sample was 26°C and after 120 seconds, it reached 28°C. Therefore, the change in temperature is also 2°C.

Next, we can use the specific heat capacity of water (4.18 J/g°C) to calculate the quantity of thermal energy absorbed by the water in each trial.

For the first trial:
Quantity of thermal energy absorbed by water = (190 g)(4.18 J/g°C)(2°C) = 1588.4 J

For the second trial:
Quantity of thermal energy absorbed by water = (170 g)(4.18 J/g°C)(2°C) = 1423.6 J

Therefore, the total quantity of thermal energy absorbed by the water in the calorimeter is 1588.4 J + 1423.6 J = 3012 J.

I hope this helps! Let me know if you have any further questions.