Question

A home 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 1

Answer : The water's temperature change will be,
`13.85^oC`

Explanation : Given,

Density of water = 0.998 g/mL

Volume of water =
`4.75* 10^2L=4.75* 10^5mL`

(conversion used : 1 L = 1000 mL)

Specific heat of water =
`4.184J/g^oC`

Heat absorbed =
`2.75* 10^4kJ=2.75* 10^7J`

(conversion used : 1 kJ = 1000 J)

First we have to determine the mass of water.

`\text{Mass of water}=\text{Density of water}* \text{Volume of water}`

`\text{Mass of water}=(0.998g/mL)* (4.75* 10^5mL)=474050g`

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

Formula used :

`Q=m* C_w* \Delta T`

where,

Q = heat absorbed by water

m = mass of water

`C_w` = specific heat of water

`\Delta T` = change in temperature

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

`2.75* 10^7J=474050g* 4.184J/g^oC* \Delta T`

`\Delta T=13.85^oC`

Therefore, the water's temperature change will be,
`13.85^oC`