In a coffee-cup calorimeter, 50.0 g hot water at 60.0 C was mixed with 50.0 g cold water at 20.0 C. If the final temperature is 36.3, what is the heat capacity of the calorimeter in J/C? Specific heat of water is 4.184 J/(g C)

2.88 J/g°C

Explanation:

Mass calorimeter (Mc) = 50g

Temperature of calorimeter (T1) = 60°C

Mass of water (Mw) = 50g

Specific heat capacity of water (Cw) = 4.184 J/g°C

Specific heat capacity of calorimeter (Cc) = ?

Temperature of water (T2) = 20°C

Final temperature of the content (T3) = 36.6°C

Assuming no heat lose from the set up

Heat loss by copper calorimeter = heat gain by the water

Mc * Cc * (T1 - T3) = Mw * Cw * (T3 - T2)

50 * Cc * (60 - 36.3) = 50 * 4.184 * (36.3 - 20)

50 * Cc * 23.7 = 209.2 * 16.3

1185Cc = 3409.96

Cc = 3409.96 / 1185

Cc = 2.877 J/g°C

Cc = 2.88 J/g°C

The specific heat capacity of the copper calorimeter is 2.88 J/g°C

it should be all of them because it asks if you included all of those in your answer

explanation:

"hydrogen bonding is the attraction of h atoms in the water molecule (slight + charge) to o atoms in a neighboring water molecule (slight - charge); it is a weak interatomic force. but it is still strong enough to the water molecules "stick" together."