How much heat does it take to warm 16.0 g of pure water from 90.0 degree C to 100.0 degree C?
The specific heat of water = 4.18 J/g. degree C

16.0 Joules
160 Joules
66.9 Joules
669 Joule

Respuesta :

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669 J

Right option is D.

Step by Step Explanation:

[tex] \sf Mass \: of \: water(m) = 16g \\ \\ \sf Specific \: heat \: of water(C) = 4.18J/g \\ \\ \sf Change \: in \: temperature ( \triangle T) = 100 - 90 = 10{ \degree}C[/tex]

  • [tex] \sf \large \red{ Heat = mC\triangle T}[/tex]

Now substituting the required values.

[tex] \sf \hookrightarrow Heat = 16 \times 4.18 \times 10 \\ \\ \sf \hookrightarrow Heat = 668.8(approx) \\ \\ \blue {\boxed { \hookrightarrow{Heat = 669J }}}[/tex]