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A coin dealer, offered a rare silver coin, suspected that it might be a counterfeit nickel copy. The dealer heated the coin, which weighed 16.5 g to 100°C in boiling water and then dropped the hot coin into 22.5 g of water at T = 15.5°C in an insulated coffee-cup, and measured the rise in temperature. If the coin was really made of silver, what would the final temperature of the water be (in °C)? (for nickel, s = 0.445 J/g°C; for silver, s = 0.233 J/g°C )

Answer :

Answer : The final temperature of the water will be, [tex]18.8^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 silver = [tex]0.233J/g^oC[/tex]

[tex]c_2[/tex] = specific heat of water = [tex]4.18J/g^oC[/tex]

[tex]m_1[/tex] = mass of silver coin = 16.5 g

[tex]m_2[/tex] = mass of water = 22.5 g

[tex]T_f[/tex] = final temperature of water = ?

[tex]T_1[/tex] = initial temperature of silver coin = [tex]100^oC[/tex]

[tex]T_2[/tex] = initial temperature of water = [tex]15.5^oC[/tex]

Now put all the given values in the above formula, we get

[tex]16.5g\times 0.233J/g^oC\times (T_f-100)^oC=-22.5g\times 4.18J/g^oC\times (T_f-15.5)^oC[/tex]

[tex]T_f=18.8^oC[/tex]

Therefore, the final temperature of the water will be, [tex]18.8^oC[/tex]

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