Chem 1 Calorimeter Question

[jinny1]

I would really appreciate if anyone could help me with this question i'm struggling with

Question A14

A calorimeter contains 95 g of water at 25°C. A 5.0 g object (at -5.0°C) is added to the water in the calorimeter and allowed to equilibrate. Calculate the final temperature of the water. The specific heat capacity of water is 4.18 J oC-1g-1 and the specific heat capacity of the object is 2.03 J oC-1g-1.

Assume no heat is lost to the calorimeter or the surroundings?

A 19.6° C
B 20.1° C
C 21.8° C
D 23.2° C
E 24.3° C

Thanks!

[Hancock]

You have to know that the sum of the Q (heat energy) in the system is equal to zero.

Therefore

Q(water) = mc*dT = 95*4.18*(Tf - 25)
Q(object) = mc*dT = 2.03*5*(Tf - (-5)) = 2.03*5*(Tf + 5)

Since sum of Q = 0

95*4.18*(Tf - 25) + 2.03*5*(Tf + 5) = 0
Solving for Tf yields E.

[jinny1]

You have to know that the sum of the Q (heat energy) in the system is equal to zero.

Therefore

Q(water) = mc*dT = 95*4.18*(Tf - 25)
Q(object) = mc*dT = 2.03*5*(Tf - (-5)) = 2.03*5*(Tf + 5)

Since sum of Q = 0

95*4.18*(Tf - 25) + 2.03*5*(Tf + 5) = 0
Solving for Tf yields E.

Thank you!

Just one thing; why is the heat energy in the system zero??

Can't that only happen if there is no change in temperature ??

[Hancock]

Heat energy is zero because the heat coming in from the hot metal is absorbed by the water. Therefore, the heat in = heat out. That's the why I look at it.