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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 )

User Melannie
by
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1 Answer

5 votes

Answer : The final temperature of the water will be,
18.8^oC

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.


q_1=-q_2


m_1* c_1* (T_f-T_1)=-m_2* c_2* (T_f-T_2)

where,


c_1 = specific heat of silver =
0.233J/g^oC


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


m_1 = mass of silver coin = 16.5 g


m_2 = mass of water = 22.5 g


T_f = final temperature of water = ?


T_1 = initial temperature of silver coin =
100^oC


T_2 = initial temperature of water =
15.5^oC

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


16.5g* 0.233J/g^oC* (T_f-100)^oC=-22.5g* 4.18J/g^oC* (T_f-15.5)^oC


T_f=18.8^oC

Therefore, the final temperature of the water will be,
18.8^oC

User Rbernabe
by
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