50g of iron with a temperature of 90#C^o# into an insulated, non conducting container with 30.0ml of water at 20#C^o#. What is the new temperature of the mixture?

C of iron=9.00 x 10#^2#
C of water=4.18x10#^2#

1 Answer
Jane
Feb 3, 2018

Here, the heat energy liberated by the iron will be taken up by the water.

Let,the final temperature of the mixture be #theta#

So, #50 *0.11(90-theta) = (30*1)1(theta-20)# (In CGS ,specific heat of iron is #0.11 Cg^-1#degree #C^-1# and weight of water = #30 cm^3*1 (gm)/(cm^3) = 30 gm#) (using #H= ms d theta#)

So,#theta= 30.84 degree C#

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