A 25-g block of iron at 175 degree C is dropped into a liter of water in an insu

Question

A 25-g block of iron at 175 degree C is dropped into a liter of water in an insulated flask at 20 degree C and 1 atm. The specific enthalpy of iron Ls given by the expression H(J/g) = 17.3T degree (C). What reference temperature was used as the basis for the enthalpy formula? Calculate the final temperature of the flask contents, assuming that the process is adiabatic, negligible evaporation of water occurs, negligible heat is transferred to the flask wall, and the specific enthalpy of liquid water at 1 atm and a given temperature is that of the saturated liquid at the same temperature. If some of the water in the flask evaporated on contact with the hot iron, would the final temperature in the flask be greater or less than the value you calculated in Part (b)? Explain your answer.

Explanation / Answer

a) Room temperature i.e., 25 oC

b) The final of the water when 25g of Iron block at 175 oC is dropped in 1 L of solution can be calculated as follows

qlost = qgain => -qiron = +qH2O

However:

q = (mass) (t) (Cp)

So:

-(mass) (t) (Cp) = (mass) (t) (Cp)

With qlost on the left side and qgain on the right side.

Substituting values into the above, we then have:

(25.0) (175 - x)(4.184) = (1000)(x - 20) (4.184)

Solve for x gives x= 23.78 oC

c) If some amount of water is evaporated form container the final temperature will be more than 23.78 oC. This is due to the fact that the heat capacity of water depends on the amount of water present in the container.

More will be the water less will be the change in temperature due to absorption of more amount of heat energy change and viceversa.

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