1. How much heat does 23.0 g of water absorb as its temperature increases from 25.4 °C to 42.8 °C? 2. A sample of unknown metal has a mass of 120.7 g. As the sample cools from 90.5 °C to 25.7 °C, it releases 7020 J of energy. What is the specific heat of the sample?

Q = 1672.84 j

c = 0.898 J/g.°C

Explanation:

1) Given data:

Mass of water = 23.0 g

Initial temperature = 25.4°C

Final temperature = 42.8° C

Heat absorbed = ?

Solution:

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

Specific heat capacity of water is 4.18 J/g°C

ΔT = 42.8°C - 25.4°C

ΔT = 17.4°C

Q = 23.0 g ×  × 4.18 J/g°C × 17.4°C

Q = 1672.84 j

2) Given data:

Mass of metal = 120.7 g

Initial temperature = 90.5°C

Final temperature = 25.7 ° C

Heat released = 7020 J

Specific heat capacity of metal = ?

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 = 25.7°C - 90.5°C

ΔT = -64.8°C

7020 J = 120.7 g ×  c ×  -64.8°C  (Negative sign shows heat is released)

7020 J = 7821.36 g.°C ×  c

c = 7020 J / 7821.36 g.°C

c = 0.898 J/g.°C

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explanation: