contestada

The specific heat capacity of lead is 0.13 J/g-°C. How much heat (in J) is required to raise the temperature of of 91 g of lead from 22°C to 37°C?

Respuesta :

Answer:

Q = 177J

Explanation:

Specific heat capacity of lead=0.13J/gc

Q=MCΔT

ΔT=T2-T1,where T1=22degrees Celsius and T2=37degree Celsius.

ΔT=37 - 22 = 15

Q = Change in energy

M = mass of substance

C= Specific heat capacity

Q = (91g) * (0.13J/gc) * (15c)= 177.45J

Approximately, Q = 177J

Otras preguntas