Question

A 325 g iron bucket contains 527 g of water, both at 21.6°C. A 855 g piece of hot steel (T = 352 degrees Celsius) is added to bucket at the same time 295 g of ice at -22.2°C is added. Determine if all the ice melts. If so, what is the final temperature of the mixture? If not, how much of the ice melted?

Answer 1

Answer:

$T = 18.63 ^oC$

Explanation:

As we know that heat given by the hot piece of iron = heat absorbed by water + ice

As we know that

$c_{iron} = 0.450 J/g ^oC$

also we have

$c_{ice} = 2050 J/kg ^oC$

now we have

$325(0.450)(T-21.6) + 0.527(4186)(T - 21.6) + 0.295(2050)(0 + 22.2) + (0.295)(335000) + 0.295(4186)(T - 0) = 855(0.450)(352 - T)$

$146.25T - 3159 + 2206T - 47650 + 13425.45 + 98825 + 1234.87T = 135432 - 384.75 T$

$3971.87 T = 73990.55$

$T = 18.63 ^oC$