Question

How would you find the temperature change in this

question?
The specific heat of lead is 0.129 JAC.
Find the amount of heat released
when 497g Pb are cooled from
37.2°C to 22.5°C.

Answer 1

To find the temperature change when cooling lead, you can use the equation Q = mcΔT, where Q is the amount of heat released, m is the mass of the substance, c is the specific heat, and ΔT is the change in temperature. In this case, the mass of lead is 497 g, the specific heat is 0.129 J/g°C, and the change in temperature is 37.2 - 22.5 = 14.7°C.

Step-by-step explanation:

To find the temperature change, we can use the equation Q = mcΔT, where Q is the amount of heat released, m is the mass of the substance, c is the specific heat, and ΔT is the change in temperature.

In this case, the mass of lead is 497 g, the specific heat is 0.129 J/g°C, and the change in temperature is 37.2 - 22.5 = 14.7°C.

Plugging these values into the equation, we get Q = (497 g)(0.129 J/g°C)(14.7°C) = 984.2565 J.