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vaieri [72.5K]
3 years ago
10

The following physical constants are for water, H2O.

Chemistry
1 answer:
Delicious77 [7]3 years ago
8 0

Answer:

Q\approx6.4~kJ

Explanation:

Quantity of heat required by 10 gram of ice initially warm it from -5°C to 0°C:

Q_1=m.C_s.\Delta T

here;

mass, m = 10 g

specific heat capacity of ice, C_s=2.09~J.g^{-1}.^{\circ}C^{-1}

change in temperature, \Delta T=(5-0)=5^{o}C

Q_1=10\times2.09\times 5

Q_1=104.5~J

Amount of heat required to melt the ice at 0°C:

Q_2=m.\Delta H_{fus}

where, \Delta H_{fus}=6020~J/mol

we know that no. of moles is = (wt. in gram) \div (molecular mass)

Q_2=\frac{10}{18} \times 6020

Q_2=3344.44~J

Now, the heat required to bring the water to 70°C from 0°C:

Q_3=m.C_L.\Delta T

specific heat of water, C_L=4.18~J/g/^oC

change in temperature, \Delta T=(70-0)=70^oC

Q_3=10\times 4.18\times 70

Q_3=2926~J

Therefore the total heat required to warm 10.0 grams of ice at -5.0°C to a temperature of 70.0°C:

Q=Q_1+Q_2+Q_3

Q=104.5+3344.44+2926

Q=6374.94~J

Q\approx6.4~kJ

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