In the sample, the mass of water is "225 g".
Given:
Heat energy,
Specific heat,
The change in temperature,
[tex]= 100-88[/tex]
[tex]= 12 ^{\circ} C[/tex]
As we know,
→ [tex]Q = mc \Delta T[/tex]
→ [tex]1.13\times 10^4=m\times 4.186\times 12[/tex]
→ [tex]m = \frac{1.13\times 10^4}{4.186\times 12}[/tex]
→ [tex]= 224.956 \ g[/tex]
or,
→ [tex]= 225 \ g[/tex]
Thus the answer above is appropriate.
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