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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?

Please show the answer, formula used and work done. Thank you!

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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

...

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