Question:
Initially a body is at a temperature of $50^{\circ}C$. If the temperature of the body is increased by $54^{\circ}F$, then its final temperature will be:
Initially a body is at a temperature of $50^{\circ}C$. If the temperature of the body is increased by $54^{\circ}F$, then its final temperature will be:
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Remember the conversion: $T(^{\circ}F) = \frac{9}{5} T(^{\circ}C) + 32$.
Updated On: Jun 6, 2026
- $80^{\circ}F$
- $90^{\circ}C$
- $176^{\circ}C$
- $176^{\circ}F$
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The Correct Option is D
Solution and Explanation
Step 1: A rise is a change.
Initial temperature $= 50^{\circ}$C and the rise $= 54^{\circ}$F. For a change of temperature, $\Delta T(^{\circ}\text{C}) = \dfrac{5}{9}\,\Delta T(^{\circ}\text{F})$.
Step 2: Rise in Celsius.
$\Delta T = \dfrac{5}{9}\times 54 = 30^{\circ}$C.
Step 3: Final value.
Final $= 50 + 30 = 80^{\circ}$C $= \dfrac{9}{5}(80)+32 = 176^{\circ}$F. \[ \boxed{176^{\circ}\text{F}} \]
Initial temperature $= 50^{\circ}$C and the rise $= 54^{\circ}$F. For a change of temperature, $\Delta T(^{\circ}\text{C}) = \dfrac{5}{9}\,\Delta T(^{\circ}\text{F})$.
Step 2: Rise in Celsius.
$\Delta T = \dfrac{5}{9}\times 54 = 30^{\circ}$C.
Step 3: Final value.
Final $= 50 + 30 = 80^{\circ}$C $= \dfrac{9}{5}(80)+32 = 176^{\circ}$F. \[ \boxed{176^{\circ}\text{F}} \]
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