If specific heat capacity of mercury is $0.033 c a l g^{- 1} {}^{0}C^{- 1}$ , how much heat is gained by $0.05 k g$ of mercury when its temperature rises from $68^{0} F$ to $313 K$ ?

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Solution

Step 1: Given data
$I n i t i a l t e m p e r a t u r e = 68^{0} F$
$F i n a l t e m p e r a t u r e = 313 K$
Specific heat capacity of mercury, $s = 0.033 c a l g^{- 1} {}^{0}C^{- 1}$
Mass of mercury, $m = 0.05 k g = 50 g$
Step 2: Convert initial and final temperature in degree celsius
The $i n i t i a l t e m p e r a t u r e = 68^{0} F$
We know that, $\frac{C}{5} = \frac{F - 32}{9}$ .
Thus, $C = [\frac{5}{9} (68 - 32)] = 20^{0} C$
$I n i t i a l t e m p e r a t u r e = 68^{0} F = 20^{0} C$
$F i n a l t e m p e r a t u r e = 313 K$
Thus, $313 K = (313 - 273) = 40^{0} C$
$F i n a l t e m p e r a t u r e = 313 K = 40^{0} C$
Change in temperature $(∆ T)$ = Final temperature - Initial temperature = $40^{0} C - 20^{0} C = 20^{0} C$
Step 3: Finding the heat gained
specific heat capacity of mercury, $s = 0.033 c a l g^{- 1} {}^{0}C^{- 1}$
Thus, Heat gained by $1 g$ of mercury for $1^{0} C$ rise $= 0.033 c a l$
Hence, Heat gained by $50 g$ of mercury for $1^{0} C$ rise $= s \times m = 0.033 \times 50 = 1.65 c a l$
Therefore, Heat gained by $50 g$ of mercury for $20^{0} C$ rise $= m s ∆ T = 1.65 \times 20 = 33 c a l$
Thus, the heat gained by $0.05 k g$ of mercury when its temperature rises from $68^{0} F$ to $313 K$ is $33 c a l$ .

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If specific heat capacity of mercury is 0.033calg-1C-10, how much heat is gained by 0.05kg of mercury when its temperature rises from 680F to 313K?

If specific heat capacity of mercury is $0.033 c a l g^{- 1} {}^{0}C^{- 1}$ , how much heat is gained by $0.05 k g$ of mercury when its temperature rises from $68^{0} F$ to $313 K$ ?