Solving Heat Capacity Problems with Specific Heat Equations

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SUMMARY

This discussion focuses on solving heat capacity problems using specific heat equations. Key calculations involve determining the specific heat of aluminum, lead, and mercury, as well as calculating the energy required to heat rhenium and ethanol. The primary equation utilized is Q=mcΔT, where Q represents heat transfer, m is mass, c is specific heat capacity, and ΔT is the change in temperature. Participants emphasize the importance of accurately applying this equation to derive specific heat values and energy requirements.

PREREQUISITES
  • Understanding of specific heat capacity concepts
  • Familiarity with the equation Q=mcΔT
  • Basic knowledge of calorimetry
  • Ability to perform unit conversions between calories and joules
NEXT STEPS
  • Practice calculating specific heat using various materials
  • Explore calorimetry experiments to measure heat transfer
  • Learn about the relationship between specific heat and phase changes
  • Investigate the specific heat capacities of different metals and liquids
USEFUL FOR

Students studying thermodynamics, chemistry enthusiasts, and educators teaching heat transfer concepts will benefit from this discussion.

WMM
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Homework Statement




1. A block of aluminum weighing 140.0 g is cooled from 98.4° C to 62.2° C with the release of 1080 cal of heat. From these data, calculate the specific heat of aluminum.

2. A total of 54.0 cal of heat are absorbed as 58.3 g of lead is heated from 12.0° C to 42.0° C. From these data, what is the specific heat of lead?

3. A 25.0 g block of rhenium metal (specific heat = 0.0329 cal/ g° C) is heated from 25.0° C to 88.2° C. How much energy is required to heat the block of rhenium?

4. A sample of mercury metal is heated from 25.5° C to 52.5° C. It took 187 cal of heat to make this temperature change. What mass of mercury was in the sample? The specific heat of mercury is 0.033 cal/ g° C

5. The specific heat of ethanol is 2.46 J/g oC. Find the heat required to raise the temperature of 193 g of ethanol from 19oC to 35oC.



Homework Equations


None I don't believe.


The Attempt at a Solution


Don't know what to star :(
 
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WMM said:

Homework Equations


None I don't believe.


Q=mcΔT will be useful.

Q = heat absorbed or released
m = mass
c = specific heat capacity
ΔT = temperature change
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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