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The specific heat capacity of lead is 0.130 J/g-K. How much heat
is required to raise the temperature of 15.0 g of lead from 22.0 °C
to 37.0 °C?

Respuesta :

Answer:

Q= 29.2 J

Explanation:

22°C=295K

37°C=310K

Q= mC dT

Q= (15.0 g) *(0.130 J/g-K)* (310 K-295K)

Q= 29.2 J

29.2 J is required to raise the temperature of 15.0 g of lead from 22.0 °C to 37.0 °C.

What is specific heat capacity?

The specific heat capacity is defined as the quantity of heat (J) absorbed per unit mass (kg) of the material when its temperature increases by 1 K (or 1 °C), and its units are J/(kg K) or J/(kg °C).

Given data:

22°C=295K

37°C=310K

Applying specific heat capacity formula:

Q= mC dT

Putting all the values in the formula:

Q= (15.0 g)  X (0.130 J/g-K) X (310 K-295K)

Q= 29.2 J

Hence, 29.2 J is required to raise the temperature of 15.0 g of lead from 22.0 °C to 37.0 °C.

Learn more about specific heat capacity here:

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