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

Answer:

The answer is 80,1 °C

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) $MassFraction_{Fe}=\frac{10g Fe}{10g Fe + 10gTi}=0.5$

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

(3) $H_{average}=0.5*0.461+0.5*0.544=0.5025\frac{J}{g-K}$

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) $-200J=20g*0.5025\frac{J}{g*K} *(T_2-373K)$

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 ( $T_1$ )

Answer 2

Answer:

80,1

Explanation:

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