thermodynamics

Question

An insulated beaker with negligible mass contains 0.310 kg of water at a temperature of 70.5°C. How many kilograms of ice at a temperature of −20.0°C must be dropped into the water to make the final temperature of the system 33.5°C?

Answer 1

Hi Shelby

Heat required by water = Mass *  Sp heat * change in temperature

 = 0.310 * 4200* (70.5 - 33.5)

= 48174 J

Which is equal to heat given by ice 

Which has three parts 

Ice from -20 C to 0 C , then melting( latent heat) and then heated from zero to 33.5 C

Point to be  noted here is that sp heat of ice is 2100 J/kg C and not 4200 J/kg C

Heat given = M * 2100 * (20 -0) + M* 336000 + M *4200 * ( 33.5 -0 ) = M ( 518700 ) = 0.09 kg

Answer 2

Hi Shelby!

 

The insulation is meant to guarantee that the negligible heat is transferred out of the beaker.  Hence, the heat to warm up the ice, melt the ice, and warm up the resulting water to 33.5 Celsius must come from the hot water.  In words, we can say that the hot water transfers heat to the ice to first warm it to 0 Celsius, then melt it (changing it from ice at 0 Celsius to water at 0 Celsius), and then warming up the resulting water from 0 Celsius to 33.5 Celsius.  The way to set this up mathematically is in the statement:

 

Q₁ + Q₂ + Q₃ + Q₄ = 0  (zero meaning no heat is transferred out of or into the beaker from the outside world)

 

where

 

Q₁ = heat needed to increase ice temperature from -20 Celsius to 0 Celsius, so it can melt

Q₂ = heat to melt the ice

Q₃ = heat to increase the temperature of the water resulting from the melting ice from 0 Celsius to 33.5 Celsius

Q₄ = heat transferred out of hot water to accomplish this (this heat value will be negative, by definition), cooling it from 70.5 Celsius to 33.5 Celsius

 

For any heat transfer which occurs without a phase change, the heat transfer is related to a change in temperature, according to:

 

Q = mcΔT

 

where

 

m = mass of material

c = specific heat of material

ΔT = T_f - T_i = change in temperature

 

If a phase change is involved, then heat transfer occurs without a change in temperature, because the heat energy goes into breaking or forming material bonds.  This is covered by the relationship:

 

Q = mL

 

where 

 

L = latent heat (of fusion, for solid-liquid transitions; or of vaporization, for liquid-gas transitions)

 

For this case, we have:

 

Q₁ = m₁c_ice(0 - (-20))  [this is the heat needed to raise the temperature of the ice so it can melt; you will need to look up the specific heat of ice, which I think is around 2090 J/kg*C)

 

Q₂ = m₁L_f-water  [this is the heat needed to melt ice at 0 Celsius to water at 0 Celsius; it depends on the latent heat of fusion of water, which I believe is 334,000 J/kg, but you should look it up

 

Q₃ = m₁c_water(33.5 - 0)  [this is the heat needed to warm up the water resulting from the melting ice to 33.5 Celsius, given that it, by definition, starts at 0 Celsius]

 

Q₄ = m₄c_water(33.5 Celsius - 70.5 Celsius)  [this is the heat given up by the hot water, which goes to doing stages 1, 2, and 3;  losing this heat, the hot water cools to the lower temperature]

 

The value m₁ is what we are trying to solve for.  Notice that it stays the same in Q₁ through Q₃, since we are talking about the same mass of ice and ice water in each of those parts.

 

Then:

 

Q₁ + Q₂ + Q₃ + Q₄ = 0

 

becomes

 

m₁c_ice(0-(-20)) + m₁L_f-water + m₁(33.5 - 0) + m₄c_water(33.5-70.5) = 0

 

with 

 

m₄ = mass of hot water = 0.310 kg

 

Since all the specific heat (c) values, and the latent heat (L) can be looked up, the only unknown in the above equation is m₁.  So, it is a bit of an algebra hike, but you can solve that equation for m₁

 

Try that out, and if you are stuck on anything or have any other questions on any part of this, just let me know.