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xeze [42]
3 years ago
15

If 75 g of metal at 100.0∘C is transferred into 80.0 mL of water at 20.0∘C initially. The water is then raised to 25∘C. What is

the specific heat of the metal?
Chemistry
1 answer:
ASHA 777 [7]3 years ago
4 0

Answer : The specific heat of the metal is, 0.297J/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 = 75 g

m_2 = mass of water  = Density\times Volume=1.00g/mL\times 80.0mL=80.0g

T_f = final temperature of mixture = 25^oC

T_1 = initial temperature of metal = 100.0.0^oC

T_2 = initial temperature of water = 20.0^oC

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

(75g)\times c_1\times (25-100.0)^oC=-[(80.0g)\times 4.18J/g^oC\times (25-20.0)^oC]

c_1=0.297J/g^oC

Therefore, the specific heat of the metal is, 0.297J/g^oC

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