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Sagot :
Answer: The specific heat of the liquid is [tex]2520J/kg^0C[/tex]
Explanation:
[tex]heat_{absorbed}=heat_{released}[/tex]
As we know that, Â
[tex]Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})[/tex]
    .................(1)
where,
q = heat absorbed or released
= mass of glass = x kg
= mass of liquid = x kg
= final temperature =
[tex]T_1[/tex] = temperature of glass = [tex]83^oC[/tex]
[tex]T_2[/tex] = temperature of liquid = [tex]43^oC[/tex]
[tex]c_1[/tex] = specific heat of glass = [tex]840J/kg^0C[/tex]
[tex]c_2[/tex] = specific heat of liquid= ?
Now put all the given values in equation (1), we get
[tex]-[(x\times 840\times (53-83)]=x\times c_2\times (53-43)[/tex]
[tex]c_2=2520J/kg^0C[/tex]
Therefore, the specific heat of the liquid is [tex]2520J/kg^0C[/tex]
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