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 = $4.75\times 10^2L=4.75\times 10^5mL$

(conversion used : 1 L = 1000 mL)

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

Heat absorbed = $2.75\times 10^4kJ=2.75\times 10^7J$

(conversion used : 1 kJ = 1000 J)

First we have to determine the mass of water.

$\text{Mass of water}=\text{Density of water}\times \text{Volume of water}$

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

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

Formula used :

$Q=m\times C_w\times \Delta T$

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:

$2.75\times 10^7J=474050g\times 4.184J/g^oC\times \Delta T$

$\Delta T=13.85^oC$

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