Complete Question
The complete question is shown on the first uploaded image
Answer:
The specific heat is [tex]c_b = 0.402 J / g \cdot ^oC[/tex]
Explanation:
From the question we are told that
The mass of the sample is [tex]m = 54.4 \ g[/tex]
The mass of the water is [tex]m_w = 150.0 \ g[/tex]
The initial temperature of the sample is [tex]T_i = 95.1 ^oC[/tex]
The initial temperature of the water is [tex]T_{w_i} = 15^oC[/tex]
The final temperature of the water is [tex]T = 17.6 ^oC[/tex]
Note the final temperature of water is equal to the final temperature of brass sample
The pressure is [tex]P =1 \ atm[/tex]
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
[tex]H_L = m * c_b * [T_i - T][/tex]
Where [tex]c_b[/tex] is the specific neat of the brass sample
The heat gained by water is mathematically evaluated as
[tex]H_g = m_w *c_w * [T_w - T ][/tex]
where [tex]c_w[/tex] is the specific heat of water which has a constant value of
[tex]c_w = 4.186 joule/gram[/tex]
So
[tex]H_L = H_g \ \equiv m* c_b * [T_i -T] = m_w * c_w * [T - T_w][/tex]
substituting values
[tex]52.4 * c_b * [95.1 - 17.6] = 150 * 4.186 * [ 17.6 - 15.0][/tex]
[tex]c_b = 0.402 J / g \cdot ^oC[/tex]
