50g of iron with a temperature of 90C^o into an insulated, non conducting container with 30.0ml of water at 20C^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^-1degree 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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