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Standard XII

Physics

Latent Heat

Assuming that...

Question

Assuming that heat released by water is entirely used to raise the temperature of calorimeter 15 $^{o}$ C to 45 $^{o}$ C, calculate the specific heat capacity of copper. (Specific heat capacity of water $= 4200 J k g^{- 1 o} C^{- 1}$

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Solution

Heat energy absorbed by calorimeter = mSt
$= 0.16 \times S \times (45 - 15) = 4.8 S$
Now, Heat given = Heat taken
$4.8 S = 2100 \Rightarrow S = \frac{2100}{4.8} = 437.5 J k g^{- 1 o} C^{- 1}$

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Assuming that heat released by water is entirely used to raise the temperature of calorimeter 15oC to 45oC, calculate the specific heat capacity of copper. (Specific heat capacity of water =4200Jkg−1oC−1

Heat energy absorbed by calorimeter = mSt=0.16×S×(45−15)=4.8SNow, Heat given = Heat taken4.8S=2100⇒S=21004.8=437.5Jkg−1oC−1

Assuming that heat released by water is entirely used to raise the temperature of calorimeter 15 $^{o}$ C to 45 $^{o}$ C, calculate the specific heat capacity of copper. (Specific heat capacity of water $= 4200 J k g^{- 1 o} C^{- 1}$

Heat energy absorbed by calorimeter = mSt
$= 0.16 \times S \times (45 - 15) = 4.8 S$
Now, Heat given = Heat taken
$4.8 S = 2100 \Rightarrow S = \frac{2100}{4.8} = 437.5 J k g^{- 1 o} C^{- 1}$