Metal X has a specific heat capacity of 0.982 Jg∙°C . If 525.0 g of Metal X increases in temperature from 32.00° C to 43.00° C, how much energy will be absorbed?

Q = 5671.05 J

Explanation:

Given data:

Mass of metal = 525 g

Initial temperature = 32.0 °C

Fina temperature = 43.0 °C

Heat absorbed by metal = ?

Solution:

Specific heat capacity:

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

specific heat capacity of metal X = 0.982 J/g.°C

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 = 43.00°C - 32.00°C

ΔT = 11.00°C

Q = 525 g × 0.982 J/g.°C  × 11.00°C

Q = 5671.05 J

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