Ahome solar energy storage unit uses 4.75 ✕ 102 l of water for storing energy. on a sunny day, the water absorbed 2.75 ✕ 104 kj of energy. how much did the water's temperature change? the density and specific heat of water are 0.998 g/ml and 4.184 j/gnaughtc. naughtc

Answer : The water's temperature change will be,

Explanation : Given,

Density of water = 0.998 g/mL

Volume of water =

(conversion used : 1 L = 1000 mL)

Specific heat of water =

Heat absorbed =

(conversion used : 1 kJ = 1000 J)

First we have to determine the mass of water.

Now we have to calculate the change in temperature of water.

Formula used :

where,

Q = heat absorbed by water

m = mass of water

= specific heat of water

= change in temperature

Now put all the given value in the above formula, we get:

Therefore, the water's temperature change will be,