Three different materials of identical mass are placed one at a time in a specia

Question

Three different materials of identical mass are placed one at a time in a special freezer that can extract energy from a material at a certain constant rate. During the cooling process, each material begins in the liquid state and ends in the solid state; Fig. 18-27 shows the temperature T versus time t. (a) For material 1, is the specific heat for the liquid state greater than or less than that for the solid state? Rank the materials according to (b) freezing-point temperature, (c) specific heat in the liquid state, (d) specific heat in the solid state, and (e) heat of fusion, all greatest first

Explanation / Answer

C=dQ/dT Since dQ/dt=constant (the rate of freezing is constant, in fact, a negative constant since you are extracting heat), we can write something neat like this: C=(dQ/dt)*(dt/dT)=dQ/dT. Since dQ/dt is constant, and dt/dT is the INVERSE OF THE SLOPE OF THE GRAPH, we know that C is proportional to the inverse of the slope. Thus C is greater if the slope is LESS steep. (a) Since the liquid state (left of the 0-slope segment) has a lesser steep, the specific heat is the greater in the liquid state than in the solid state. (b) From the graph, you can read right off the T-axis: from greatest freezing point to lowest freezing point: 1,2,3 (c) C in the liquid state: from greatest to least (that is, lowest slope to highest slope): 1,3,2 (d) C in the solid state: from greatest to least (that is, lowest slope to highest slope): 1,2,3 (e) The energy required for fusion is just integral((dQ/dt)dt), which is a constant times the change in time. Thus, the longer the 0-slope segment is, the more energy was required to go into fusion. From greatest to least (that is, longest 0-slope segment to shortest): 2,3,1

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