Question

when 27 g of water absorbs 1,5000 joules of heat energy the temperature of the water is raised to 57.7 what is the initial temperature of the water

Answer 1

Answer:

The initial temperature of the water is -75.08 K.

Explanation:

Given that,

Mass of water, m = 27 g

Heat absorbed, Q = 1,5000 J

Final temperature of water, T₂ = 57.7 K

The specific heat of water is 4.184 J/​g-K

We know that,

$Q=mc\Delta T\\\\Q=mc(T_2-T_1)\\\\\dfrac{Q}{mc}=(T_2-T_1)\\\\T_1=T_2-\dfrac{Q}{mc}$

Put all the values,

$T_1=57.7-\dfrac{15000}{27\times 4.184 }\\\\=-75.08\ K$

So, the initial temperature of the water is -75.08 K.