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Standard XII

Physics

Calorimetry

Calculate the...

Question

Calculate the mass of ice needed to cool $150$ g of water contained in a calorimeter of mass $50$ g at $32^{o}$ C such that the final temperature is $5^{o}$ C.
Specific heat capacity of calorimeter $= 0.4 J / g^{o} C$
Specific heat capacity of water $= 4.2 J / g^{o} C$
Latent heat capacity of ice $= 330 J / g$ .

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Solution

Mass of water $m_{w} = 150 g$
Specific heat capacity of water $S_{w} = 4.2 J / g^{o} C$
Initial temperature $T_{i} = 32^{o} C$
Final temperature $T_{f} = 5^{o} C$
Mass of calorimeter $m_{c} = 50 g$
Specific heat capacity of calorimeter $S_{c} = 0.4 J / g^{o} C$
Heat released $Q = (m_{w} S_{w} + m_{s} S_{s}) (T_{i} - T_{f})$
$∴$ $Q = (150 \times 4.2 + 50 \times 0.4) (32 - 5) = 17550 J$
Mass of ice required $m_{i} = \frac{Q}{L}$
where $L = 330 J / g$
$∴$ $m_{i} = \frac{17550}{330} = 53.2 g$

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Calculate the mass of ice needed to cool 150g of water contained in a calorimeter of mass 50g at 32oC such that the final temperature is 5oC.Specific heat capacity of calorimeter=0.4J/goCSpecific heat capacity of water=4.2J/goCLatent heat capacity of ice=330J/g.