Iron metal has a specific heat of 0.449j/g°c. how much heat is transferred to a 5.00 a piece of iron, initially at 20.0°c, when it is placed in a beaker of boiling water at 1 atm?

use q= mcat formula
for chemistry

Q = 44.9 j

Explanation:

Mass of iron = 5.0 g

Change in temperature = 20 °C

Specific heat of iron = 0.449 j/g.°C

Heat transferred = ?

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

Solution:

Q = m.c. ΔT

Q = 5 g × 0.449 j/g.°C ×  20 °C

Q = 44.9 j

d: the voltage of the secondary coil is less than that of the primary coil.

im not suree sorry if wrong