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?Answer:__________________Work: (required for an A)

Answer 1

Step-by-step explanation

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

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

From the table,

For the first iron heated to 60°C

`\begin{gathered} T_1=60,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,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