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3 years ago

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A metal sample weighing 129.00 grams and at a temperature of 97.8 degrees Celsius was placed in 45.00 grams of water in a calori

meter at 20.4 degrees Celsius. At equilibrium the temperature of the water and metal was 39.6 degrees Celsius. Calculate the specific heat of the metal. The specific heat of the water is 4.184 J/g/C

1 answer:

Nitella [24]3 years ago

4 0

Answer : The specific heat of metal is $0.481J/g^oC$ .

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 = $4.184J/g^oC$

$m_1$ = mass of metal = 129.00 g

$m_2$ = mass of water = 45.00 g

$T_f$ = final temperature = $39.6^oC$

$T_1$ = initial temperature of metal = $97.8^oC$

$T_2$ = initial temperature of water = $20.4^oC$

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

$129.00g\times c_1\times (39.6-97.8)^oC=-45.00g\times 4.184J/g^oC\times (39.6-20.4)^oC$

$c_1=0.481J/g^oC$

Therefore, the specific heat of metal is $0.481J/g^oC$ .

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