finding energy required to warm to melting point, temperature once it has reached its melting point, and how much energy is required to melt after its reached i

Question

Gallium, Ga: 69.723 g/mole

Solid specific heat, Csolid 0.371 J/g C

Enthalpy of Fusion, 80.2 J/g

melting point temperature, Tmelt 29.76 C

Liquid specific heat, Cliquid 0.371 J/g C

Suppose you cast 20.0 g spoon out of pure Gallium, cool it to an initial temperature of 0.00 C and use it to stir a 250 mL cup of coffee that starts at a temperature of 95.00 C. Assume that the coffee has a heat capacity identical to water (4.184 J/g C) and a density of 1.00 g/mL

a)Assuming that the entire spoon is heated uniformly, how much energy is required to warm the Gallium to its melting point?

b)what is the temperature of the coffee after the Gallium has been warmed to its melting point?

c)how much energy is required to melt the Gallium after it has reached its melting point?

d.what is the temperature of the coffee after this energy transfer

Answer 1

Once the spoon is placed in the coffee, heat will flow from the higher to lower temperature object (by the 2nd law of thermodynamics).  Assuming the two are in thermal isolation -- which we have to assume, or we do not have enough information -- they will come to the same temperature, which is the point where heat will stop flowing; and all the heat transferred from the higher temperature body is transferred to the lower temperature body.

 

The amount of heat transferred is determined in two ways.  If there is no phase change, heat transferred (Q) is related to a change in temperature by:

 

[1]  Q = mcΔT

 

where

 

m = mass of material

c = specific heat (the amount of heat transfer needed to change one unit mass of material by one degree

ΔT = change in temperature

 

If there IS a phase change, then heat transfer is not associated with a change in temperature, but rather with the formation or breaking of bonds.  The relationship is:

 

[2]  Q = mH

 

where H = enthalpy (of fusion, for solid-liquid boundary crossing, or of vaporization, for liquid-gas;  there is also one for solid-gas conversion (sublimation), though it is used less often)

 

This whole process could be dealt with in a single equation (with adequate preparation), but his problem breaks it up into parts.

 

a)  How much energy [thermal energy transfer; i.e. heat] is needed to warm the spoon to its melting point?

 

Use Equation [1] for this, with the given information and knowing ΔT = T_f - T_ii.  Since this stage ends with the spoon just reaching its melting temperature, but not yet melting, T_f = 29.76 ^oC, and T_i = 0.00 ^oC.  Confirm that all the units match between specific heat and the other quantities

 

Using Equation [1] in this case, I obtained:  Q_a = 220.8 J

 

b)  The coffee at least starts off by cooling, giving up an amount of heat equal to what the spoon absorbs in Part a).  We can use this to figure our a change in coffee's temperature, starting (at least) with Equation [1] again.  We use the result from Part a) as the amount of heat, rewrite ΔT again as Tf-Ti, and solve for T_f, as

 

Q = m_cofC_cofc(T_f - Ti)   [T_f is the only unknown]

 

Two notes:

-- since this heat is transferred out of the coffee, we have to put in Q for the coffee as negative, by definition

-- we do have to take the additional step of solving for m_cof, since we are given a volume and a density.

 

With all that together, I obtained:  T_f = 94.79 ^oC

 

(this just confirms, by the way, that the coffee is not brought anywhere near a phase change by this heat transfer, because the amount of heat need to change the temperature of the spoon is pretty small)

 

c)  The heat transfer needed to change the spoon's phase by melting is calculated using Equation [2] and given information.  With that, I calculated Q_melt = 1,604 J  (note that it take a LOT more energy to change phase than to change temperature)

 

d)  The energy to melt the spoon also comes from the coffee, and -- starting with the assumption that the coffee, once again, does not change phase -- this heat energy can be calculated by Equation [1], with the same notes as in Part b).  With that, I obtained T_f2 = 93.26 ^oC.  (once again, this confirms that there is not nearly enough heat transfer to get the coffee to a phase change point).

 

If you have any more questions about the algebra details here, I would be happy to talk further in an online session, so just let me know!