Question

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.

Answer 1

Complete Question

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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

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

    $c_b = 0.402 J / g \cdot ^oC$