Question

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

Answer 1

`{\huge{\fcolorbox{yellow}{red}{\orange{\boxed{\boxed{\boxed{\boxed{\underbrace{\overbrace{\mathfrak{\pink{\fcolorbox{green}{blue}{Answer}}}}}}}}}}}}}`

669 J

Right option is D.

Step by Step Step-by-step explanation:

`\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`

• 
  `\sf \large \red{ Heat = mC\triangle T}`

Now substituting the required values.

`\sf \hookrightarrow Heat = 16 * 4.18 * 10 \\ \\ \sf \hookrightarrow Heat = 668.8(approx) \\ \\ \blue {\boxed { \hookrightarrow{Heat = 669J }}}`