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anonymous · Answer

what's the question?

anonymous · Answer

Did I do it correctly? I got:

mcΔT(metal) = mcΔT(water)
27.778 x c x (100.5 - 38.9) = 25 x 4.18 x (38.9 - 25.3)
27.776 x c x 61.6 = 25 x 4.18 x 13.6
1711.0016 x c = 1421.2
C= 1.2

And the question was:

Calculate the energy change (q) of the surroundings (water) using the enthalpy equation

qwater = m × c × ΔT.

We can assume that the specific heat capacity of water is 4.18 J / (g × °C) and the density of water is 1.00 g/mL.

The water has absorbed the heat of the metal. So, qwater = qmetal.

Data:
Measured Mass of Metal: 27.776 g
Distilled Water Volume: 25 mL
Distilled Water Temperature: 25.3 mL
Temperature of Metal: 100.5 C
Temperature of Mixture: 38.9 C

anonymous · Answer

first of all find heat loss by metal and heat gain by water .if heat gain by water is equal to heat loss by water then there is no energy  change of surrounding.if heat loss by metal is greater than heat gain by water then remaining amount of heat loss by metal will go to surrounding and will increase the energy of surrounding.if heat loss by metal is less than the heat gain by the water then water will absorb  extra heat  from the surrounding i.e heat of surrounding will decrease in this case.....

anonymous · Answer

@xmelodymusic ,are you there???

anonymous · Answer

in the first line of above post,correction:
 if heat gain by water is equal to heat loss by metal then there is no energy  change of surrounding

anonymous · Answer

@xmelodymusic :rply me...do you understand that post.

anonymous · Answer

I'm here, sorry, walked away from the computer. So, my answer was wrong?

anonymous · Answer

yes this is not correct process...

anonymous · Answer

Distilled Water Temperature: 25.3 mL....correct it in your question 25.3C not ml

anonymous · Answer

@xmelodymusic :hey in your question ,what is the specific heat  capacity of metal.it is missing in your question...

anonymous · Answer

Okay, sorry! Just a second, let me re-look it.

anonymous · Answer

the specific heat capacity of water is 4.18 J / (g × °C)

anonymous · Answer

that's all I have

anonymous · Answer

this time,only post the question once again..