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Drupady [299]
3 years ago
5

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
Chemistry
1 answer:
Sav [38]3 years ago
8 0

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

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