Question

How much heat is released as 5.00G of PB cool from 75.0° C to 25.0° C

Answer 1

To calculate the heat released as 5.00g of lead cools from 75.0°C to 25.0°C, use the formula Q = mc\(\Delta\)T with the specific heat of lead. The calculation yields 32.0 J of heat released.

Step-by-step explanation:

The amount of heat released as a substance cools can be calculated using the formula Q = mc\(\Delta\)T where Q is the heat energy, m is the mass, c is the specific heat capacity, and \(\Delta\)T is the change in temperature. For lead (Pb), the specific heat capacity is approximately 0.128 J/g°C. To find the heat released as 5.00g of Pb cools from 75.0°C to 25.0°C, simply plug the values into the formula. The temperature change \(\Delta\)T is 75.0°C - 25.0°C = 50.0°C. The calculation becomes Q = (5.00 g)(0.128 J/g°C)(50.0°C).

Q = (5.00 g)(6.4 J/°C)

Q = 32.0 J of heat released.

Answer 2

32.25 J

Step-by-step explanation:

Heat, H = ?

Mass, m = 5g = 5 / 1000 = 0.005 Kg

Initial Temperature = 75.0° C

Final Temperature = 25.0° C

Temperature change, ΔT = Final - Initial = 50° C

Specific heat capacity of lead, C = 129 (J/kg°C)

The relationship between these quantities is given by the equation

H = mCΔT

H = 0.005 * 129 * 50

H = 32.25 J