Question

Given the data, find the: a) Temp. change of metal (±0.1 °C) b) Heat absorbed by water (Joule) c) Heat lost by metal (Joule) d) Specific heat of the metal (J/g °C)

Answer 1

(a) 10(±0.1 °C)

Step-by-step 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 =

`((22.2+22.4+23.7))/(3)(0.1\degree C)=((68.3))/(3)(±0.1\degree C)=22.8(±0.1\degree C)`

The average final temperature, T₂ =

`(31.3+34.0+33.2)/(3)(\pm0.1\degree C)=(98.5)/(3)(\pm0.1\degree C)=32.8(\pm0.1\degree C)`

Therefore, the temp. change of metal (±0.1 °C) ΔT = T₂ - T₁

`ΔT=32.8-22.8=10(\pm0.1\degree C)`

(b) Heat absorbed by water (Joule).

The formula to calculate the heat absorbed by water in joule is:

`\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}`

The average mass of water is

`m_w=(9.007+8.811+8.826)/(3)=(26.644)/(3)=8.881\text{ }\pm0.001g`

The specific heat of water is