Answer:
The specific heat of the sample [tex]\mathbf{c_3 = 1011.056 \ J/kg.K}[/tex]
Explanation:
Given that:
mass of an unknown sample [tex]m_3[/tex] = 0.0850
temperature of the unknown sample [tex]t_{unknown}[/tex] = 100° C
initial temperature of the calorimeter can = 19° C
mass of copper [tex]m_1[/tex] = 0.150 kg
mass of water [tex]m_2[/tex]= 0.20 kg
the final temperature of the calorimeter can = 26.1° C
The objective is to compute the specific heat of the sample.
By applying the principle of conservation of energy
[tex]Q = mc \Delta T[/tex]
where;
[tex]Q_1 +Q_2 +Q_3 = 0[/tex]
i.e
[tex]m_1 c_1 \Delta T_1 +m_2 c_2 \Delta T_2+m_3 c_3 \Delta T_3 =0[/tex]
the specific heat capacities of water and copper are 4.18 × 10³ J/kg.K and 0.39 × 10³ J/kg.K respectively
the specific heat of the sample [tex]c_3[/tex] can be computed by making [tex]c_3[/tex] the subject of the above formula:
i.e
[tex]c_3 = \dfrac{m_1 c_1 \Delta T_1 +m_2 c_2 \Delta T_2}{m_3 c_3 \Delta T_3}[/tex]
[tex]c_3 = \dfrac{ 0.150 \times 0.39 \times 10^3 \times (26.1 -19) + 0.20 \times 4.18 \times 10^3 \times (26.1 -19) }{0.0850 \times (100-26.1 )}[/tex]
[tex]c_3 = \dfrac{ 0.150 \times 0.39 \times 10^3 \times (7.1) + 0.20 \times 4.18 \times 10^3 \times (7.1) }{0.0850 \times (73.9)}[/tex]
[tex]c_3 = \dfrac{415.35 + 5935.6 }{6.2815}[/tex]
[tex]c_3 = \dfrac{415.35 + 5935.6 }{6.2815}[/tex]
[tex]c_3 = \dfrac{6350.95}{6.2815}[/tex]
[tex]\mathbf{c_3 = 1011.056 \ J/kg.K}[/tex]
The specific heat of the sample [tex]\mathbf{c_3 = 1011.056 \ J/kg.K}[/tex]
