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

true

explanation:

answers: to make sure that all of the broken pieces of tablet were added to the water, to allow for fair comparison of all three trials, with different particle sizes to avoid experimental error