Calculate the amount of heat lost by 23 g of water when its temperature drops from 68.0°C to 31.0°C. The specific heat of water is 4.18 J/(g°C).

3557J is the heat lost by the water

Explanation:

Using the equation:

Q = m*C*ΔT

Where Q is change in heat

m is mass of substance (23g of water)

C is specific heat of the substance (4.18J/g°C)

And ΔT is change in temperature (68.0°C - 31.0°C = 37.0°C)

we can find the amount of heat lost as follows:

Q = 23g*4.18J/g°C*37°C

Q =

3557J is the heat lost by the water

b) mutualism

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