Calorimetry problem and temperature

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SUMMARY

The discussion revolves around a calorimetry problem where a student attempts to determine the specific heat capacity of an unknown metal. The experiment involves a 100g metal heated to 75°C, placed in a 70g copper calorimeter with 200g of water at 20°C, resulting in a final temperature of 25°C. The calculations indicate that the specific heat capacity of the metal is approximately 864 J/kg°C, but the student is uncertain about the accuracy of this value compared to the provided options, which include 4200 J/kg*K and 900 J/kg*K, suggesting that the correct answer may be 900 J/kg*K, potentially indicating aluminum.

PREREQUISITES
  • Understanding of specific heat capacity and its calculation
  • Familiarity with calorimetry principles
  • Knowledge of heat transfer equations
  • Basic understanding of units of measurement in physics (Joules, kg, °C)
NEXT STEPS
  • Review the principles of calorimetry and heat transfer
  • Learn how to calculate specific heat capacity using heat transfer equations
  • Study the properties of common metals, focusing on their specific heat capacities
  • Explore the concept of thermal equilibrium in calorimetry experiments
USEFUL FOR

This discussion is beneficial for students studying physics, particularly those focusing on thermodynamics and calorimetry, as well as educators seeking to clarify concepts related to specific heat capacity and heat transfer in experiments.

crosbykins
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Homework Statement


A student attempts to identify a metal by measuring its specific heat capacity. 100g of the metal is heated to 75°C and then transferred to a 70g copper calorimeter containing 200g of water at 20°C. The temperature of the final mixture is 25°C.


Homework Equations



heat lost by sample = heat gained by water

The Attempt at a Solution


Given metal: Given water:
m = .1kg m = .2kg
c = ? c = 4.2e3 J/kg°C
t1 = 75°C t1 = 20°C
t2 = 25°C t2 = 25°C
c = - [.2kg * 4.2e3 J/kg°C]/ *25°C - 20°C/
.1kg 25°C - 75°C
= 840 J/kg°C
ok well..this is wrong...does the fact that the calorimeter is 70g copper affect the answer
 
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Do you how much heat is needed to change the temperature of 1 kg of water by 1 degree Celsius. It is 4186 Joule. So How much heat is needed to change the temperature of 200 g of water by 1 degree Celsius ? It is 5 time less heat than 1 Kg of water. But you have changed the temperature of water by 5 degree Celsius. So total heat needed to change the temperature of 200 g of water by 5 degree Celsius is:

(4186 * 5)/5

= 4186 Joule.

Now copper also needed some heat to raise its temperature by 5 degree Celsius.

We need 387 Joule of heat to raise the temperature of 1 kg of copper by 1 degree Celsius.

So to raise the temperature of 70 g copper by 5 degree Celsius is ...?

Now add both the heat and find what is the total heat lost by the metal and due to this lost how much is the change in the temperature. From there you can find its specific heat.
 
It may be nearly equal to 864.29 J/kg per degree Celsius. Is I'm right? I'm not very good with Physics so I apologize if I've mislead you on any step.
 
snshusat161 said:
Do you how much heat is needed to change the temperature of 1 kg of water by 1 degree Celsius. It is 4186 Joule. So How much heat is needed to change the temperature of 200 g of water by 1 degree Celsius ? It is 5 time less heat than 1 Kg of water. But you have changed the temperature of water by 5 degree Celsius. So total heat needed to change the temperature of 200 g of water by 5 degree Celsius is:

(4186 * 5)/5

= 4186 Joule.

Now copper also needed some heat to raise its temperature by 5 degree Celsius.

We need 387 Joule of heat to raise the temperature of 1 kg of copper by 1 degree Celsius.

So to raise the temperature of 70 g copper by 5 degree Celsius is ...?

Now add both the heat and find what is the total heat lost by the metal and due to this lost how much is the change in the temperature. From there you can find its specific heat.
-so copper's heat gained is (387J *5)/14 =138J...so the heat gained by both is 4324J...so that's the heat lost by the sample, so the sample's specific heat capacity is...864J/Kg*C...but that's no the right answer..b/c the choices i have to choose from for the answer are 4200J/Kg*K, 900J/Kg * K, 440J/kg *, 390, 370, 130...
 
Tell the correct answer. Is it 900 J/ kg *K. Is it Aluminum.
 

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