A 345 g chunk of gold (c= .126 J/gK) at 235 C is dropped into 826 g of water at 20.5 C. What will the final temperature of the gold be after the system reaches equilibrium?
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Let T = final temperature in ° C.
Heat lost by gold = heat gained by water
=> 345 * 0.126 * (235 - T) = 826 * 4.186 * (T - 20.5)
=> 43.47 (235 - T) = 3457.636 (T - 20.5)
=> (3457.636 + 43.47) T = (3457.636 * 20.5 + 43.47 * 235)
=> 3501.106 T = 81096.988
=> T = 23.16° C.
Heat lost by gold = heat gained by water
=> 345 * 0.126 * (235 - T) = 826 * 4.186 * (T - 20.5)
=> 43.47 (235 - T) = 3457.636 (T - 20.5)
=> (3457.636 + 43.47) T = (3457.636 * 20.5 + 43.47 * 235)
=> 3501.106 T = 81096.988
=> T = 23.16° C.