Calculations:
We are given :
• mass of water = ?
• ΔT, =( Tfinal - T initial ) = ( 40 -30 )= 10 °C
,• Q ,= 250J
,• Specific heat capacity(,c,) = 4.2J/g•°C
We will the formula :
[tex]Q\text{ = m}\cdot c\cdot\text{ }\Delta T\text{ }[/tex]Therefore ;
[tex]\text{Mass = Q/(c}\cdot\text{ }\Delta T)\text{ }[/tex]Replacing the given parameters, we get that mass =
[tex]\begin{gathered} \text{mass }=\frac{250J}{4.2J\cdot\text{ 10}\degree C}\cdot g\degree C \\ \text{ = 5.95 grams } \end{gathered}[/tex]• This means that the mass required = 5.95 grams.
