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Iteru [2.4K]
3 years ago
11

A sample of brass is put into a calorimeter (see sketch at right) that contains of water. The brass sample starts off at and the

temperature of the water starts off at . When the temperature of the water stops changing it's . The pressure remains constant at . Calculate the specific heat capacity of brass according to this experiment. Be sure your answer is rounded to significant digits.

Chemistry
1 answer:
Ierofanga [76]3 years ago
4 0

Complete Question

The complete question is shown on the first uploaded image

Answer:

The specific heat is  c_b  = 0.402 J / g \cdot ^oC

Explanation:

From the question we are told that

    The mass of  the sample is  m = 54.4 \ g

     The mass of the water is  m_w = 150.0 \ g

     The initial temperature of the sample is  T_i  = 95.1 ^oC

     The initial temperature of the water is  T_{w_i} =  15^oC

     The final temperature of the water is  T  =  17.6 ^oC

Note the final temperature of water is equal to the final temperature of brass sample

    The pressure is  P =1 \ atm

Generally for according to the law of energy conservation

    The heat lost by sample  =  The heat gain by water

   

The heat lost by brass sample is  mathematically evaluated as    

          H_L  =  m * c_b  *  [T_i - T]

Where c_b is the specific neat of the brass sample

The heat gained  by water is  mathematically evaluated as          

        H_g  = m_w *c_w * [T_w - T ]

where c_w is the specific heat of water which has a constant value of  

     c_w =  4.186  joule/gram

So

    H_L  =  H_g \ \equiv m* c_b  * [T_i -T] =  m_w * c_w * [T - T_w]

substituting values

    52.4 * c_b  * [95.1  - 17.6] =  150 * 4.186 * [ 17.6 - 15.0]

    c_b  = 0.402 J / g \cdot ^oC

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