Question

Calculate the final temperature that results from mixing 245 grams of cobalt, specific heat 0.446 J/g0C, at 1420C with 106 grams of water at 24.80C.

Answer 1

Heat needed to raise water to 100$^0$ C:

$0.106\times4.18\times(100-24.8)=.331Kj$

heat needed to change water to steam$= 2264.76kJ/kg\times0.106= 240kj$

Heat needed to raise steam from 100 to Tf.

$0.106\times2kj/kg\times(Tf-100)=???$

Now, heat given off by cooling Cobalt:

$0.245\times0.446\times(tf-1460)$

sums of heats = 0

$0.245\times0.446\times(tf-1460)-240-0.106\times2kj/kg\times(Tf-100)-0.331=0$

Now solve for $T_f$ and easily get your answer.