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gregori [183]
3 years ago
7

. 125g of water has an initial temperature of 25.6°C, and is heated by 50.0g of a metal

Chemistry
1 answer:
Eduardwww [97]3 years ago
4 0

Answer : The specific heat of the metal is, 0.451J/g^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\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

c_1 = specific heat of metal = ?

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

m_1 = mass of metal = 50.0 g

m_2 = mass of water  = 125 g

T_f = final temperature of mixture = 29.3^oC

T_1 = initial temperature of metal = 115.0^oC

T_2 = initial temperature of water = 25.6^oC

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

(50.0g)\times c_1\times (29.3-115.0)^oC=-[(125g)\times 4.18J/g^oC\times (29.3-25.6)^oC]

c_1=0.451J/g^oC

Therefore, the specific heat of the metal is, 0.451J/g^oC

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