When a heat engine is used as a refrigerator to lower the temperature of an object, the colder the object the more work that is needed to cool it further to the same extent. (a) suppose that the refrigerator is an ideal heat engine and that it extracts a quantity of heat |dq| from the cold source (the object being cooled) at temperature tc. the work done on the engine is |dw| and as a result heat (|dq|+|dw|) is discarded into the hot sink at temperature t h . explain how the second law requires that, for the process to be allowed, the following relation must apply: |dq|tc=|dq|+|dw|th. (b) suppose that the heat capacity of the object being cooled is c (which can be assumed to be independent of temperature) so that the heat transfer for a change in temperature dtc is dq=cdtc. substitute this relation into the expression derived in (a) and then integrate between tc=ti and tc=tl to give the following expression for the work needed to cool the object from ti to tl as w=ctb∣∣lntitl∣∣−|c(tl−ti)|. (c) use this result to calculate the work needed to lower the temperature of 250 g of water from 293 k to 273 k, assuming that the hot reservoir is at 293 k. (cpm(h2o(l))=75.3jk−1mol−1). (d) when the temperature of liquid water reaches 273 k it will freeze to ice, an exothermic process. calculate the work needed to transfer the associated heat to the hot sink, assuming that the water remains at 273 k (the standard enthalpy of fusion of h2o is 6.01 kjmol−1 at the normal freezing point). (e) hence calculate the total work needed to freeze the 250 g of liquid water to ice at 273 k. how long will this take if the refrigerator operates at 100 w?