Solving Heat Capacity Problems with Specific Heat Equations

  • Thread starter WMM
  • Start date
In summary, these conversations involve using specific heat capacity to calculate the amount of energy required to heat or cool a substance. In the first conversation, the specific heat of aluminum is calculated by cooling a 140.0 g block from 98.4° C to 62.2° C with the release of 1080 cal of heat. In the second conversation, the specific heat of lead is found by heating a 58.3 g sample from 12.0° C to 42.0° C with the absorption of 54.0 cal of heat. The third conversation involves calculating the energy needed to heat a 25.0 g block of rhenium from 25.0° C to 88.
  • #1
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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  • #2
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
 

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