Heat energy is given to $80 g$ alcohol (sp heat capacity $2200 J k g^{- 1} K^{- 1}$ ) when its temperature rises by $20 K$ . If the same energy is given to the $200 g$ mercury of a specific heat capacity $140 J k g^{- 1} K^{- 1}$ . Now the rise in temperature is to be found out?

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Solution

Step 1: Calculating heat energy given to alcohol
Heat energy given to $\frac{80}{1000} kg$ of alcohol $= m C Δ t$
$= \frac{80}{1000} \times 2200 \times 20 = 3520 J$
Step 2: Calculating the change in temperature of mercury,
Energy given to $\frac{200}{1000} k g$ of mercury $= (\frac{200}{1000}) \times 140 \times Δ t$
$(\frac{200}{1000}) \times 140 \times Δ t = 3520 ∆ t = \frac{3520}{28} ∆ t = 125.7 K$
Hence, the rise in the temperature is $125.7 K .$

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Heat energy is given to 80g alcohol (sp heat capacity 2200Jkg-1K-1) when its temperature rises by 20K. If the same energy is given to the 200gmercury of a specific heat capacity 140Jkg-1K-1. Now the rise in temperature is to be found out?

Heat energy is given to $80 g$ alcohol (sp heat capacity $2200 J k g^{- 1} K^{- 1}$ ) when its temperature rises by $20 K$ . If the same energy is given to the $200 g$ mercury of a specific heat capacity $140 J k g^{- 1} K^{- 1}$ . Now the rise in temperature is to be found out?