Chemistry, thermachemistry

Question

Exactly 16.5 mL of water at 32.0 °C is added to a hot iron skillet. All of the water is converted to steam at 100.0°C. The mass of the skillet is 1.45 kg. What is the change in temperature of the skillet?

Answer 1

I will show you how to work this, but you do the calculations.  You need to work this in 2 parts.  

 

Part 1.

 

Water is heated from 32°C to 100°C.

 

m_w = mass of water (multiply the 16.5 mL volume by the density of water to get mass)

c_w = specific heat of water (look it up in your chemistry book)

ΔT_w = temperature change of the water = (100 - 32)°C = ?

 

First heat transfer to the water:

 

Q₁ = m_wc_wΔT_w

 

Part 2.

 

Water is converted to steam at a constant temperature of 100°C.

 

l_v = latent heat of vaporization of water (look it up in your chemistry book)

 

Second heat transfer to the water:

 

Q₂ = m_wl_v

 

Total heat transfer to the water is

 

Q = Q₁ + Q₂

 

The skillet cools down, so you need to find the heat transfer into the skillet (which is a negative quantity, since the temperature of the skillet will drop), and balance it with the heat transfer into the water.

 

m = mass of iron = 1.45 kg

c = specific heat of iron (look it up in your chemistry book)

ΔT = temperature change of the iron (this is what you want to solve for)

 

Note that ΔT is negative, since the iron is cooling down.

 

mcΔT = heat transfer into the iron (a negative number, since heat will flow out of the hot iron and into the cooler water)

 

The net heat transfer into the system must be zero, so

 

Q₁ + Q₂ + mcΔT = 0 ⇒

 

ΔT = -(Q₁ + Q₂) / (mc)

 

You can see that ΔT will come out negative, as you would expect.  Plug in the numbers, and be careful witht the units!