Thank you for visiting A metal sample weighing 129 00 grams and at a temperature of 97 8 degrees Celsius was placed in 45 00 grams of water in. This page is designed to guide you through key points and clear explanations related to the topic at hand. We aim to make your learning experience smooth, insightful, and informative. Dive in and discover the answers you're looking for!
Answer :
Answer : The specific heat of metal is [tex]0.481J/g^oC[/tex].
Explanation :
In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.
[tex]q_1=-q_2[/tex]
[tex]m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)[/tex]
where,
[tex]c_1[/tex] = specific heat of metal = ?
[tex]c_2[/tex] = specific heat of water = [tex]4.184J/g^oC[/tex]
[tex]m_1[/tex] = mass of metal = 129.00 g
[tex]m_2[/tex] = mass of water = 45.00 g
[tex]T_f[/tex] = final temperature = [tex]39.6^oC[/tex]
[tex]T_1[/tex] = initial temperature of metal = [tex]97.8^oC[/tex]
[tex]T_2[/tex] = initial temperature of water = [tex]20.4^oC[/tex]
Now put all the given values in the above formula, we get
[tex]129.00g\times c_1\times (39.6-97.8)^oC=-45.00g\times 4.184J/g^oC\times (39.6-20.4)^oC[/tex]
[tex]c_1=0.481J/g^oC[/tex]
Therefore, the specific heat of metal is [tex]0.481J/g^oC[/tex].
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Rewritten by : Jeany
Final answer:
To calculate the specific heat of the metal, the heat transfer equation q=mcΔT is used. By setting the heat lost by the metal equal to the heat gained by the water, and substituting the known values into the equation, we can solve for the specific heat of the metal.
Explanation:
The specific heat of a metal can be calculated by using the concept of heat transfer, where heat lost by the metal is equal to the heat gained by the water in a calorimetry experiment. The equation is q = mcΔT, where q is the heat transfer, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. For the water, the heat gained can be calculated as qwater = mwatercwaterΔTwater. For the metal, the heat lost is qmetal = mmetalcmetalΔTmetal. Given that the heat lost by the metal equals the heat gained by the water, the equation can be set up to solve for the specific heat of the metal, cmetal = (qwater / (mmetalΔTmetal)). We know the following: mwater = 45.00 g, cwater = 4.184 J/g/C, ΔTwater = final temperature - initial temperature of water, mmetal = 129.00 g, ΔTmetal = initial temperature of metal - final temperature. By substituting these values into the equation, we can find the specific heat of the metal.
