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Hi

A piece of copper of mass 20.0g at 100oC is placed in a vessel of negliglible heat capacity but containing 50.7 g of water at 22oC. Calculate the final temperature of the water.

The answer is 25oC but i dont know how you get this. The question pertains to a section on heat capacity. The heat capcities of water and copper are 4.184 and 0.38 J/gK respectively

I would presume heat would be exchanged until the copper and the water had the same temperature but i wouldn't know how to calculate when exactly this happened except by trial and error

.thanks

A piece of copper of mass 20.0g at 100oC is placed in a vessel of negliglible heat capacity but containing 50.7 g of water at 22oC. Calculate the final temperature of the water.

The answer is 25oC but i dont know how you get this. The question pertains to a section on heat capacity. The heat capcities of water and copper are 4.184 and 0.38 J/gK respectively

I would presume heat would be exchanged until the copper and the water had the same temperature but i wouldn't know how to calculate when exactly this happened except by trial and error

.thanks

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heat1 = mass of water x initial water temperature x water hcapacity

heat1 = mass of copper x initial copper temperature x coppes hcapacity

Now solve equation, where the only unknown value is final temperature

totalheat = final temperature x mass of water x water hcapacity + final temperature x mass of copper x copper hcapacity

Enjoy :)

*Note:

m=Mass

c=Specific heat capacity

Q= Temperature change

x = Theta (unknown/Temperature)

Temperature released by copper = Temperature taken in by water

mass * specific heat capacity * (Original Temp - Final Temp) = mass * Specific heat capacity * (Final temp - Original Temp)

20g * 0.38J/gK * (100C - x) = 4.184J/gK * 50.7 * (x - 22C)

7.6J/K * (100C - x) = 212.1288 J/K * (x - 22C)

Next Step: Expansion of equation:

760 JC/K - 7.6x J/K = 212.1288x J/K - 4666.8336 JC/K

Movement of equation:

5426.8336 JC/K = 219.7288x J/K

5426.8336 C = 219.7288x (I just cancelled the units since they were common)

x = 5426.8336 / 219.7288

x = 24.69787119C

x ~ 25C (Rounded up to 2 significant figures)

Hope this is useful! (Anyway, I am a Singapore Secondary 4 Student taking the GCSE Exams this year.)

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Start your 7-day free trialHow can i solve that? I don't have the final temperature of the Cu or the water?

In the equation:

mcQ(Copper)=mcQ(Water)

Mass of Copper, water respectively: 20g , 50.7g

Heat capacities of Cu, H20 respectively: 0.38 J/gK , 4.184 J/gK

Temperature of Cu, H20 respectively: 100C, 22C

The common problem is substituting the temperature inside the equation.

Analysis: By comparing the temperature of water and copper, you can immediately conclude that copper will lose heat to the water, that is a fact. After you can conclude which material gains and loses heat/energy, you would then be able to apply the formula.

So, temperature loss by copper is calculated as (Original - Final) or (100 - x),

because final temperature of copper would definitely be lower than 100C.

(Take x to be an unknown because you do not know the temperature yet).

And, temperature gained by water is calculated as (Final - Original) or (x-22),

because final temperature of water would be higher than 22C after copper releases heat into the water.

From this amount of information given, you should be able to solve this question already.

Math / Science

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q = C * m(g) * D(T)

q = 4.184 * 50.7 * 3 = 636.3864 J

If the copper lost this much energy its temperature would decrease by

636.3864 = 0.38 * 20 * D(T)

D(T) = 83.74

Temp of Cu :

100-83.74 = 16.26oC

This doesnt make sense