Answer
The best estimate for the missing data point is 37 °C
Explanation:
Heat lost by copper = heat gain by water, i.e
[tex]\begin{gathered} -Q=+Q \\ -m_cc_c(T_3-T_1)=m_wc_w(T_3-T_2) \end{gathered}[/tex]From the table.
For the first copper heated to 60 °C
[tex]\begin{gathered} T_1=60,\text{ }T_3=25 \\ -m_cc_c(25-60)=m_wc_w(25_{}-T_2) \\ -m_cc_c(-35)=m_wc_w(25_{}-T_2) \\ m_cc_c(35)=m_wc_w(25_{}-T_2)----i \end{gathered}[/tex]For the second copper heated to 75 °C
[tex]\begin{gathered} T_1=75,\text{ }T_3=31 \\ -m_cc_c(31-75)=m_wc_w(31_{}-T_2) \\ -m_cc_c(-44)=m_wc_w(31_{}-T_2) \\ m_cc_c(44)=m_wc_w(31_{}-T_2)----ii \end{gathered}[/tex]For the third copper heated to 90 °C
[tex]\begin{gathered} T_1=90,\text{ Let }T_3=x \\ -m_cc_c(x-90)=m_wc_w(x_{}-T_2) \\ m_cc_c(-x+90)=m_wc_w(x_{}-T_2) \\ m_cc_c(90-x)=m_wc_w(x_{}-T_2)----iii \end{gathered}[/tex]Since the mass of the copper blocks are the same and amount of the water are equal, then divide (i) by (ii)
[tex]\begin{gathered} \frac{m_cc_c\mleft(35\mright)}{m_cc_c(44)}=\frac{m_wc_w(25-T_2)}{m_wc_w(31-T_2)} \\ \\ \frac{35}{44}=\frac{(25-T_2)}{(31-T_2)} \\ \text{Cross multiply} \\ 44(25-T_2)=35(31-T_2) \\ 1100-44T_2=1085-35T_2 \\ \text{Combine the like terms} \\ 1100-1085=44T_2-35T_2 \\ 15=9T_2 \\ To\text{ get }T_2,\text{ divide both sides by 9} \\ \frac{15}{9}=\frac{9T_2}{9} \\ T_2=1.67^oC \end{gathered}[/tex]To get x in (iii) which is the missing equilibrium temperature for the third copper heated to 90 °C, substitute T₂ = 1.67 °C into (i) and (ii) and divide (ii) by (iii)
[tex]\begin{gathered} \frac{m_cc_c(44)}{m_cc_c(90-x)}=\frac{m_wc_w\mleft(31_{}-1.67\mright)}{m_wc_w\mleft(x_{}-1.67\mright)} \\ \\ \frac{44}{90-x}=\frac{29.33}{x-1.67} \\ \text{Cross multiply} \\ 44(x-1.67)=29.33(90-x) \\ 44x-73.48=2639.7-29.33x \\ \text{Combine the like terms} \\ 44x+29.33x=2639.7+73.48 \\ 73.33x=2713.18 \\ \text{Divide both sides by 73.33} \\ \frac{73.33x}{73.33}=\frac{2713.18}{73.18} \\ x=37^0C \end{gathered}[/tex]Therefore, the best estimate for the missing data point is 37 °C.
