Posted by bbsdn on Sunday, January 27, 2013 at 2:35pm.
You can do a quick calculation to see if Ag will reach the melting point. That will be
q = 40 x 0.235 x (Tf-Ti) = 40 x 0.235 x (961-20) = about 9,000 J and you don't have that many joules; i.e., only 1490 J available. Therefore, substitute into the equation and calculate Tf.
1490J = [mass Ag x specific Ag x (Tf-Ti)]
1490 = [40 x 0.235 x (Tf-20)] = about 180 C or so. You can do it more accurately.
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