A 21.5 g piece of iron at 100.0∘C is dropped into 132 g of water at 20.0∘C. What is the final temperature of the system, in degrees Celsius, if the specific heat of iron is 0.449Jg∘C? Round your answer to one decimal place. Use 4.184Jg∘C for the specific heat of water.

T = 21.37 °C

Explanation:

In this case, we need to use the following expression:

Heat of iron = Heat of water  (1)

This is because they are both related. It's supposed that the iron piece which is in a higher temperature, will lose some heat after it's dropped into water. Therefore, the heat that the iron loses, would be the heat that the water gains.

So, the general expression for heat in this case would be:

Q = m dT H  (2)

And we know that dT is the difference in the temperatures of the iron at first and at the end, for one side. On the other side, it would be the difference of the final temperature and the innitial temperature of water.

Now, if we use (1) and (2) together we will have the following:

m₁ dT₁ H₁ = m₂ dT₂ H₂   (3)

1 is iron, and 2 is water.

Now dT₁ and dT₂ would be:

dT₁ = Ti - T

dT₂ = T - Tw

Now replacing in (3) we have:

m₁(Ti - T) H₁ = m₂(T - Tw) H₂  

21.5 . (100 - T) . 0.449 = 132 . (T - 20) . 4.184

9.6535(100 - T) = 552.288(T - 20)

965.35 - 9.6535T = 552.288T - 11045.76

11045.76 + 965.35 = 552.288T + 9.6535T

12011.11 = 561.9415T

T = 12011.11 / 561.9415

T = 21.37 °C

i'm pretty sure its b i might be wrong though!

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3