Question

Equilibrium temperature of water and copper blockSample original temperatureEquilibrium temperature60°C25°C75°C31°C90°C????????3 copper blocks of equal masses are heated to different temperatures. These blocks are then dropped into 3 different beakers filled with equal amounts of water. What is the best estimate for the missing data point? I NEED THE ANSWER FOR (90°C)???????? WHAT'S THE Equilibrium temperature!

Answer 1

The best estimate for the missing data point is 37 °C

Step-by-step explanation:

Heat lost by copper = heat gain by water, i.e

`\begin{gathered} -Q=+Q \\ -m_cc_c(T_3-T_1)=m_wc_w(T_3-T_2) \end{gathered}`

From the table.

For the first copper heated to 60 °C

`\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}`

For the second copper heated to 75 °C

`\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}`

For the third copper heated to 90 °C

`\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}`

Since the mass of the copper blocks are the same and amount of the water are equal, then divide (i) by (ii)

`\begin{gathered} (m_cc_c\mleft(35\mright))/(m_cc_c(44))=(m_wc_w(25-T_2))/(m_wc_w(31-T_2)) \\ \\ (35)/(44)=((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} \\ (15)/(9)=(9T_2)/(9) \\ T_2=1.67^oC \end{gathered}`

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)

`\begin{gathered} (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)} \\ \\ (44)/(90-x)=(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} \\ (73.33x)/(73.33)=(2713.18)/(73.18) \\ x=37^0C \end{gathered}`

Therefore, the best estimate for the missing data point is 37 °C.