1,200 cal of heat energy is added to a liquid with a specific heat of 0.57 cal/g°C. If the temperature increases from 20.°C to 33°C, what is the mass of the liquid?

The mass of the liquid = 162 g

Explanation:

As we know ,  

Q = mc (ΔT)

where Q = Heat energy of the liquid

c = specific heat of the liquid

ΔT = change in temperature

m = mass of the liquid

As given , the temperature increases from 20.°C to 33°C,

⇒ ΔT = 33 - 20 = 13°C

Also given , c = 0.57 cal/g°C

                   Q = 1200 cal

∴ we get

1200 = (m)(0.57)(13)

⇒ 1200 = m(7.41)

⇒m = = 161.943 ≈ 162 g

So, we get

The mass of the liquid = 162 g

b) mutualism

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