Answer
(a) 10(±0.1 °C)
Explanation
(a) Temp. change of metal (±0.1 °C)
To find the temp. change of metal (±0.1 °C), you need to first determine the averages of the initial and final temperatures of the water:
The average initial temperature, T₁ =
[tex]\frac{(22.2+22.4+23.7)}{3}(0.1\degree C)=\frac{(68.3)}{3}(±0.1\degree C)=22.8(±0.1\degree C)[/tex]The average final temperature, T₂ =
[tex]\frac{31.3+34.0+33.2}{3}(\pm0.1\degree C)=\frac{98.5}{3}(\pm0.1\degree C)=32.8(\pm0.1\degree C)[/tex]Therefore, the temp. change of metal (±0.1 °C) ΔT = T₂ - T₁
[tex]ΔT=32.8-22.8=10(\pm0.1\degree C)[/tex](b) Heat absorbed by water (Joule).
The formula to calculate the heat absorbed by water in joule is:
[tex]\begin{gathered} q=m_wc_w\Delta T \\ Where; \\ q\text{ }is\text{ }the\text{ }heat\text{ }absorbed\text{ }by\text{ }water \\ m_w\text{ }is\text{ }the\text{ }mass\text{ }of\text{ }water \\ c_w\text{ }is\text{ }the\text{ }specific\text{ }heat\text{ }of\text{ }water \\ \Delta T\text{ }is\text{ }the\text{ }temp.\text{ }change \end{gathered}[/tex]The average mass of water is
[tex]m_w=\frac{9.007+8.811+8.826}{3}=\frac{26.644}{3}=8.881\text{ }\pm0.001g[/tex]The specific heat of water is
