A22.5 g sample of a metal is heated to 65.00 °c and added to 25.00 g of water (c = 4.184 j/g °c) in a calorimeter. the water temperature rises from 25.55 °c to 35.25 °c. what is the specific heat capacity of the metal?

The specific heat capacity of the metal is 1.143 J/g°C

Explanation:

A typical excersise of calorimetry:

Q = m . C . ΔT

ΔT = Final T° - Initial T°

m = mass

In this case, the heat released by the metal is gained by the water to rise its temperature.

Qmetal = Qwater

(We consider that metal was at the same T° of water)

22.5 g  . C . (65°C - 25.55°C) = 25 g .  4.184 J/g°C . (35.25°C - 25.55°C)

22.5 g  . C . 39.45°C = 25g . 4.184 J/g°C . 9.7°C

887.625 g.°C . C = 1014.62J

C = 1014.62J / 887.625 g.°C

C = 1.143 J/g°C

when [oh-] is 10-5  m, and [h3o+] is 10-9  m, the poh is 5 and ph is 9. again in both cases the sum of ph and poh is 14. however, the hydrogen ion concentration is not always going to be equal to exactly 1 x 10 raised to a negative number. for example, we skipped over the value of 2.0 x 10-7.