Question

1. The specific heat capacity of iron is 0.461 J g–1 K–1 and that of titanium is 0.544 J g–1 K–1. A sample consisting of a mixture of 10.0 g Fe and 10.0 g Ti at 100.0 ºC loses 200. J of heat to the environment. What is the final temperature of the sample?

Answer 1

80,1

Step-by-step explanation:

write the equation out and see what is what and then do the math

Answer 2

The answer is 80,1 °C

Step-by-step explanation:

Let´s start from the mass of the sample and the heat capacities:

First of all, we must calculate an average heat capacity. That's because we have a mixture and it is unknown the heat capacity of the whole sample.

The way we should do this calculation is as follows:

(1)
`H_(average)=Mass Fraction_(first component)* H_(first component)+MassFraction_(secondcomponent)*H_(second component)`

For example, the mass fraction of Fe is simply:

(2)
`MassFraction_(Fe)=(10g Fe)/(10g Fe + 10gTi)=0.5`

If you combine the equations (1) and (2) you have:

(3)
`H_(average)=0.5*0.461+0.5*0.544=0.5025(J)/(g-K)`

Once calculated the average heat capacity we can solve the problem taking into account the corresponding equation:

(4)
`Q=m*H_(average)*(T_2-T_1)`

Remember that:

Q: Heat gained or lost

m: Mass of the sample you want to analize

`H_(average)` : The value obtained in equation (3)

`T_2`: Final temperature of the sample

`T_1`: Initial temperature of the sample

Now we must replace the problem data in equation (4)

Take into account:

• Heat gained in a system have a positive value
• Heat lost in a system have a negative value

• In this problem the sample loses 200 J, for this reason
  `Q=-200J`

• The mass of the whole sample is: 10g of Fe + 10g of Ti = 20g of sample
• The temperatures must be in absolute units of temperature (these are: rankine or kelvin)
• The initial temperature of the system is 100°C or 373K

Now we are ready to use equation (4):

(5)
`-200J=20g*0.5025(J)/(g*K) *(T_2-373K)`

It is clear that the unknown in equation (5) is
`T_2`

The next step is to calculate
`T_2`. Don't forget the signs; these are important.

Key concept: Since the system is loosing heat, the final temperature of the system (
`T_2`) should be lower than the initial temperature (
`T_1` )