Chemistry: If 14.5 kj of heat were added to 485 g of liquid water, how much would its temperature increase?

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If 14.5 kj of heat were added to 485 g of liquid water,

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02.01.2022

The increase in temperature is 7.14°C

Explanation:

To calculate the change in temperature, we use the equation:

$q=mc\Delta T$

where,

q = heat absorbed = 14.5 kJ = 14500 J (Conversion factor: 1 kJ = 1000 J)

m = mass of water = 485 g g

c = specific heat capacity of water = 4.184 J/g.°C

$\Delta T$ = change in temperature = ?

Putting values in above equation, we get:

$14500J=485g\times 4.184J/g.^oC\times \Delta T\\\\\Delta T=7.14^oC$

Hence, the increase in temperature is 7.14°C

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