How to Calculate the Specific Heat Capacity of an Unknown Metal

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To calculate the specific heat capacity of an unknown metal, the heat gained by water must equal the heat lost by the metal. The formula used is ΔE = m * ΔT * c, where ΔE represents the energy change. In this case, the heat gained by the water can be calculated using its mass and temperature change, leading to a value of 4445.5 Joules. Substituting this into the rearranged specific heat capacity formula yields a calculated value of 4.3. The discussion emphasizes understanding the relationship between heat transfer and the specific heat capacity of both the metal and water.
Verbally
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Homework Statement




How can you calculate the specific heat capacity of an unknown metal? In my question the metal with a mass of 50g and a temperature of 200 degrees celsius was placed in 125 g of water with an initial temperature of 20 degrees celsius. After the simulation and the metal was placed in the water, the temperature of the water rose from 20 to 28.35. How can I find the specific heat capacity of the unknown metal? Then we have to research what metal corresponds to the resulting heat capacity.

Homework Equations



I know you have to use the original equation of ΔE=m*ΔT*c and rearrange it to isolate "c", thus resulting in the equation of c=ΔE/m*ΔT. But how do you calculate ΔE?


The Attempt at a Solution



What I did first is try to find ΔE:

= m x (T2-T1)
= 125 x (28.35-20)
= 125 x 8.35
= 1043.75 Calories = 4445.5 Joules

Then you can substitute that into c=ΔE/m*ΔT

c = 4445.5J / 125 x 8.35
= 4445.5 / 1043.75
= 4.3

I know I went wrong somewhere, could someone please explain to me where I did? I've been stuggling on this question forever!

Thanks.
 
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Heat lost by metal equals heat gained by water.
 
Borek said:
Heat lost by metal equals heat gained by water.

Ok, but how do you find the heat gained by water?
 
You could find the heat change in the water by reasoning from the definition that 1 calorie raises 1 g of water by 1 degree Celsius. That and the data you have let you get the answer through dimensional analysis.

If you want to be more mechanical (and, arguably, more efficient), you also have an equation that relates the change of temperature, mass, specific heat and change of heat content.

Do you see it?
 
Verbally said:
Ok, but how do you find the heat gained by water?

From the same m*ΔT*c - just for water.
 
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