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On a new scale of temperature (which is linear) and called the `W` scale. The freezing and boiling points of water are `39^(@)W` and `239^(@)W` respectively. What will be the temperature on the new scale, corresponding to a temperature of `39^(@)C` on the Celsius scale? |
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Answer» In general, whenever we are to go from any known scale to any unknow scale, then we follow the equation `=(("Temperature on one scale")-("LFP for known scale"))/((UFP-LFP)_("known"))` `=(("Temperature on other scale")-("LFP for unknown scale"))/((UFP-LFP)_("unknown"))` `rArr " "(39-0)/(100-0)=(t-39)/(239-39)rArrt=117^(@)W` |
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