The final temperature of the given water will be "25.71°C".
The given values are:
Mass of water,
Specific heat,
Heat absorbed,
Initial temperature,
As we know,
The change in temperature,
→ [tex]\Delta = T_f-T_i[/tex]
Now,
→ [tex]q = ms \Delta T[/tex]
or,
→ [tex]q = ms(T_f - T_i)[/tex]
By substituting the given values, we get
→ [tex]950=150.1\times 4.18\times (T_f -T_i)[/tex]
→ [tex]T_f-T_i = \frac{950}{150.1\times 4.18}[/tex]
→ [tex]T_f-24.2 = \frac{950}{627.418}[/tex]
→ [tex]T_f -24.2 = 1.51[/tex]
→ [tex]T_f = 1.51+24.2[/tex]
→ [tex]=25.71^{\circ} C[/tex]
Thus the solution above is right.
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