Respuesta :
Explanation:
The given data is as follows.
Mass of copper calorimeter, ([tex]m_{cu}[/tex]) = 0.1 kg
specific heat of copper calorimeter, ([tex]c_{cu}[/tex]) = 390 J/kg K
mass of water, ([tex]m_{w}[/tex]) = 0.160 kg
specific heat of water, ([tex]c_{w}[/tex]) = 4190 J/kg K
mass of ice, ([tex]m_i[/tex]) = 0.018 kg
latent heat of ice = [tex]334 \times 103 J/kg[/tex]
mass of lead, ([tex]m_{Pb}[/tex]) = 0.75 kg
specific heat of lead, ([tex]c_{Pb}[/tex]) = 130 J/kg K
Heat lost by the lead is
Q = [tex]m_{pb} \times C_{Pb} \times (255 - T)[/tex]
= [tex]0.75 \times 130 J/kg K \times (255 - T)[/tex]
= 97.5 (255 - T)
Now, heat gained by calorimeter is calculated as follows.
Q = [tex](m_{w} + m_{ice}) \times c_{w} \times T + m_{Cu} \times C_{Cu} \times T + m_{ice} \times L_{f}[/tex]
= [tex](0.160 + 0.018)(4190) T + (0.1)(390 J/kg K) T + (0.018 kg )(334 \times 103)[/tex]
Hence, heat loss by lead is equal to heat gained by calorimeter .
So, 97.5 (255 - T ) = 745.82 T + 39 T + 6012
18850.5 = 882.32 T
T = [tex]21.36^{o}C[/tex]
Thus, we can conclude that the value of final temperature is [tex]21.36^{o}C[/tex].
