Help needed with this h=mcat question please!!!! (1 Viewer)

[flowerp]

Hi every1,

I need help with the following questions:

A copper cube with a mass of 100 grams is heated in a boiling water bath to 100.0 Celsius. The cube is removed from the bath and trasnferred to a beaker of 200mL water at 25 celsius. What is the temp change of this water after the cu block is added?


Thanks

 

[Atrium]

So this question requires you to recognize that the copper cube has reached 100 degrees celsius since it's equilibrated with the boiling water.

When the cube is transferred to a beaker of water at 25 degrees celsius the total heat of the system will redistribute and re-equilibrate.

What this means is that the heat that was held in the copper cube will transfer directly to water, and that the total heat of the system combined is a constant (law of conservation of energy).

Therefore, Q(Cu) = Q(Water)
Q(Cu) = (0.1)(0.386)(100 - x)
Q(Water) = (0.2)(4.186)(x - 25)
Solve for x.

 

[flowerp]

Aahh, I see! Thank you so much!!!!!

So this question requires you to recognize that the copper cube has reached 100 degrees celsius since it's equilibrated with the boiling water.

When the cube is transferred to a beaker of water at 25 degrees celsius the total heat of the system will redistribute and re-equilibrate.

What this means is that the heat that was held in the copper cube will transfer directly to water, and that the total heat of the system combined is a constant (law of conservation of energy).

Therefore, Q(Cu) = Q(Water)
Q(Cu) = (0.1)(0.386)(100 - x)
Q(Water) = (0.2)(4.186)(x - 25)
Solve for x.