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Q.

A glass beaker contains 200g of carbonated water initially at 20°C How much ice should be added to obtain the final temperature of 0°C with all ice melted, if the initial temperature of ice is -10°C Neglect heat capacity of glass [ $Take, C_{water} = 4190 J / k g^{0} C,$ $C_{i c e} = 2100 J / k g^{0} C, L_{F} = 3.34 \times 10^{5} J / k g]$

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a

47g

b

76g

c

200g

d

22g

answer is A.

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Detailed Solution

Heat lost by carbonated water = Heat gained by ice
Let m₁ = mass of ice melted and m_w = initial mass of carbonated water.
$m_{w} c_{w} (20 - 0) = m_{1} L_{f} + m_{i} (c_{i}) [0 - (- 10)]$
$\Rightarrow 200 \times 10^{- 3} \times 4190 \times 20 = m_{1} (3.34 \times 10^{5} + 2100 \times 10) \Rightarrow m_{1} = 47 \times 10^{- 3} k g o r m_{1} = 47 g$

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