In an experiment, 21.5 g of metal was heated to 98.0°c and then quickly transferred to 150.0 g of water in a calorimeter. the initial temperature of the water was 21.5°c, and the final temperature after the addition of the metal was 32.5°c. assume the calorimeter behaves ideally and does not absorb or release heat.
what is the value of the specific heat capacity (in j/g•°c) of the metal? j/g•°c

The value of the specific heat capacity is 4.89 J/g°C

Explanation:

First of all, let's make the equation.

Assume the calorimeter behaves ideally and does not absorb or release heat, this phrase explains that all the heat released the metal, was gained by the water so:

Qmetal = metal mass . C metal . ΔT

Q water = water mass . C H2O . ΔT

Q metal = Q water

21.5 g . C metal (32.5 °C - 98°C ) = 150 g . 4.18 J/g°C (32.5°C - 21.5°C)

21.5 g . C metal . -65.5 °C = 150g . 4.18 J/g°C . 11°C

21.5 g . C metal . -65.5 °C = 6897 J

C metal =  6897 J /21.5 g . 65.5°C

C metal = 4.89 J/g°C

Notice that specifc heat is a constant. It can be negative. NEVER

We use the absolute value. |x|

in scientific notation, the value of the  speed of light in a vacuum

  will be written as 2.995 * 10^8. when this value is represented in the form of

  a * 10^b, then, a = 2.995 and b = 8.

the value of the speed of light in a vacuum given above can also be rounded up to approximately 300000000. in this case, the scientific notation will be written as 3 *10^8.