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Finding the equilibrium temperature of the tin-water mixture

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A pan of water is heated from 23 C to 78 C. What is the change in its temperature on the kelvin and fahrenheit scales?

A 0.225 kg sample of tin initially at 97.5 C is dropped into 0.115 kg of water initially at 10.0 C. If the specific heat capacity of tin is 230 J/kg *C (its a dot before "C" but im not exactly sure how to put it), what is the final equilibrium temperature of the tin-water mixture?

https://brainmass.com/physics/equilibrium/equilibrium-temperature-tin-water-mixture-13332

SOLUTION This solution is FREE courtesy of BrainMass!

To convert celcius to farenheit, we have the formula,
F = (9/5)C + 32

where, F is the farenheit temperature and C is the temp in degree Celcius.

And to convert celcius to kelvin scale,
K = C + 273

23 degree Celcius = (9/5)23 + 32 Farenheit = 73.4 F
and 23 C = 23 + 273 = 296 K

Similarly,
78C = (9/5)78 + 32 = 172.4 F

and 78 C = 78 + 273 = 351 K

Now, change in temperature in degree clecius = 78 - 23 = 55C

In farenheit = 172.4 - 73.4 = 99
In kelvin, = 351 - 296 = 55K

Note that the temp change in celcius scale and kelvin scale are the same.

We know that heat transferred, q = m * c *(T2 - T1)
where, m is the mass, c is the heat capacity and T1 and T2 are the initial and final temp of the substance.

Since energy must be conserved,heat lost by tin will be equal to the heat absorbed by water.
Here, m(tin)= 0.225 Kg, m(water)= 0.11Kg
c(tin)=230J/Kg/c , c(water) = 4190 J/Kg/C
T2 is the equilibrium temperature.
Hence we can write,
0.225*230*(T2 - 97.5) = -0.115 * 4190 * (T2 - 10)
51.75 * T2 - 5045.625 = -481.85 * T2 + 4818.5
533.6 * T2 = 9864.125
T2 = 18.386 C

This is the final equilibrium temperature of the tin-water mixture.

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