Question

If 14.5 kJ of heat were added to 485 g of liquid water, How much would its temperature increase?

Answer 1

Answer: 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