A) \[18.2{}^\circ C\]
B) \[22{}^\circ C\]
C) \[20.2{}^\circ C\]
D) \[24.2{}^\circ C\]
E) \[20.8{}^\circ C\]
Correct Answer: C
Solution :
Ist case: \[m{{s}_{A}}(t-{{t}_{A}})=m{{s}_{B}}({{t}_{B}}-t)\] \[{{s}_{A}}(16-12)={{s}_{B}}(19-16)\] \[4{{s}_{A}}=3{{s}_{B}}\] lInd case: \[m{{s}_{B}}(t-{{t}_{B}})=m{{s}_{C}}({{t}_{C}}-t)\] \[{{s}_{B}}(23-19)={{s}_{C}}\,(28-23)\] \[4{{s}_{B}}=5{{s}_{C}}\] \[3{{s}_{B}}=\frac{15}{4}{{s}_{C}}\] \[\therefore \] \[4{{S}_{A}}=3{{s}_{B}}=\frac{15}{4}{{s}_{C}}\] \[\Rightarrow \] \[16{{s}_{A}}=12{{s}_{B}}=15{{s}_{C}}=k\] \[{{s}_{A}}:{{s}_{B}}:{{s}_{C}}=\frac{1}{16}:\frac{1}{12}:\frac{1}{15}\] \[{{s}_{A}}=\frac{k}{16},{{s}_{c}}=\frac{k}{15}\] When A and C are mixed \[m{{s}_{A}}(t-{{t}_{A}})=m{{s}_{C}}({{t}_{C}}-t)\] \[\frac{k}{16}(t-12)=\frac{k}{15}(28-t)\] \[15t-180=448-16t\] \[31t=628\] \[\Rightarrow \] \[t=20.2{}^\circ C\]
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