Respuesta :
Answer:
The specific heat of the unknown sample is 1822.14 J/kg.k
Explanation:
Given;
mass of aluminum calorimeter, [tex]M_c[/tex] = 100 g
mass of water, [tex]M_w[/tex] = 250 g
stabilizing temperature of water-calorimeter, ΔT = 10.0°C
mass of copper, [tex]M_c_u[/tex] = 75 g
initial temperature of copper, [tex]T_{cu}[/tex] = 60.0°C
mass of unknown sample, [tex]M_u[/tex] = 70.0 g
initial temperature of unknown sample, [tex]T_u[/tex] = 100°C.
The final temperature of the entire system, t = 20.0°C
Apply the principle of conservation of energy;
energy used to heat water and calorimeter is equal to energy released by copper and unknown sample.
[tex]Q = M_{cu}C_{cu} \delta T_{cu} + M__{u}C{_u} \delta T_u[/tex]
where;
Q is energy used to heat water and calorimeter
[tex]C_c_u[/tex] is the specific heat capacity of copper
[tex]C_u[/tex] is the specific heat capacity of unknown sample
Make [tex]C_u[/tex] subject of the formula;
[tex]C_u = \frac{Q-M_c_u C_c_u \delta T_c_u}{M_u \delta T_u} \\\\C_u = \frac{(C_wM_w +C_cM_c)\delta T-M_c_u C_c_u \delta T_c_u}{M_u \delta T _u} \\\\C_u = \frac{(4186*0.25 +900*0.1)10-0.075* 387 *40}{0.07* 80} \\\\C_u = \frac{11365 -1161}{5.6} \\\\C_u = 1822.14 \ J/kg.k[/tex]
Therefore, the specific heat of the unknown sample is 1822.14 J/kg.k
The specific heat of the unknown sample is;
c_u = 1823 J/Kg.k
We are given;
Mass of Aluminum calorimeter; m_c = 100 g = 0.1 kg
Mass of water; m_w = 250 g = 0.25 kg
Initial temperature of Calorimeter and water; T_c = T_w = 10°C = 283 K
Mass of Copper; m_cu = 75 g = 0.075 kg
Initial temperature of Copper; T_cu = 60°C = 333 K
Mass of unknown sample; m_u = 70 g = 0.07 kg
Initial temperature of unknown substance = 100°C = 373 K
Final temperature of system; T_f = 20°C = 293 K
Formula for quantity of heat is;
Q = mcΔt
where;
m is mass
c is specific heat capacity
Δt is change in temperature;
For the calorimeter and water , we have;
Q_cw = (m_w*c_w + m_c*c_c)Δt
Specific heat capacity of water is; c_w = 4186 J/Kg.K
Specific heat capacity of aluminium is 900 J/Kg.K
Thus;
Q_cw = ((0.25 * 4186) + (0.1 * 900))(293 - 283)
Q_cw = 11365 J
For the unknown sample and the piece of copper;
Q_cu,u = (m_cu*c_cu*Δt) + (m_u*c_u*Δt)
specific heat capacity of copper; c_cu = 385 J/Kg.K
Thus;
Q_cu,u = (0.075*385*(333 - 293)) + (0.07*c_u*(373 - 293))
Q_cu,u = 1155 + 5.6c_u
From conservation of energy principle;
Q_cw = Q_cu,u
Thus;
11365 = 1155 + 4.2c_u
c_u = (11365 - 1155)/5.6
c_u = 1823 J/Kg.k
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