Question

How much hear must be obsorberd by 375.0 grams of water to raise its temperature from 10.0c to 35.0c? The specific heat capacity of water is 4.186 J/gC.
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Answer 1

Answer:

The heat absorbed by water is 39243.75 J.

Explanation:

Given data:

Mass of water = 375.0 g

Heat absorbed by water= ?

Initial temperature = 10.0°C

Final temperature = 35.0 °C

The specific heat capacity of water = 4.186 j/g.°C

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = T2 - T1

ΔT = 35°C - 10°C

ΔT = 25°C

Q = m.c. ΔT

Q = 375.0 g× 4.186 j/g °C × 25°C

Q = 39243.75 J

The heat absorbed by water is 39243.75 J.