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Sagot :
Answer: The specific heat of the metal in [tex]1.34J/g^0C[/tex]
Explanation:
[tex]Q_{absorbed}=Q_{released}[/tex]
As we know that,
[tex]Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})[/tex]
[tex]m_1\times c\times (T_{final}-T_1)=-[m_2\times c\times (T_{final}-T_2)][/tex]
where,
[tex]m_1[/tex] = mass of metal = 31.7 g
[tex]m_2[/tex] = mass of water = 100.0 g
[tex]T_{final}[/tex] = final temperature = [tex]24.6^oC[/tex]
[tex]T_1[/tex] = temperature of metal = [tex]96.5^oC[/tex]
[tex]T_2[/tex] = temperature of water = [tex]17.3^oC[/tex]
[tex]c_1[/tex] = specific heat of metal= ?
[tex]c_2[/tex] = specific heat of water = [tex]4.184J/g^0C[/tex]
Now put all the given values in equation (1), we get
[tex]m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)][/tex]
[tex]-(31.7\times c_1\times (24.6-96.5)^0C)=(100.0\times 4.184\times (24.6-17.3)][/tex]
[tex]c_1=1.34J/g^0C[/tex]
Therefore, the specific heat of the metal in [tex]1.34J/g^0C[/tex]
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