Respuesta :
Answer : The energy required to heat of 1.50 kg iron is, [tex]1.5\times 10^4J[/tex]
Explanation :
Formula used :
[tex]Q=m\times c\times \Delta T[/tex]
or,
[tex]Q=m\times c\times (T_2-T_1)[/tex]
where,
Q = heat required = ?
m = mass of iron = 1.50 kg = 1500 g
c = specific heat of iron = [tex]0.450J/g^oC[/tex]
[tex]T_1[/tex] = initial temperature = [tex]-7.8^oC[/tex]
[tex]T_2[/tex] = final temperature = [tex]15.0^oC[/tex]
Now put all the given value in the above formula, we get:
[tex]Q=1500g\times 0.450J/g^oC\times (15.0-(-7.8))^oC[/tex]
[tex]Q=15390J=1.5\times 10^4J[/tex]
Therefore, the energy required to heat of 1.50 kg iron is, [tex]1.5\times 10^4J[/tex]
Answer:
There is 15.4 kJ of energy required to heat 1.50 kg iron.
Explanation:
Step 1: Data given
Mass of iron = 1.50 kg = 1500 grams
Initial temperature iron = -7.8 °C
Final temperature iron = 15.0 °C
Specific heat capacity of iron = 0.450 J/g*K = 0.450 J/g°C
Note: when heating iron from -7.8 °C to 15.0 °C, the iron does not change to another phase.
Step 2: Calculate the energy required
q = m*c*ΔT
⇒ with Q = the heat transfer ( in Joules)
⇒ with m= the mass of iron = 1.5 kg = 1500 grams
⇒ with c= the specific heat capacity of ir = 0.450 J/g°C
⇒ with ΔT = The change in temperature = T2 - T1 = 15.0 + 7.8 = 22.8 °C
q = 1500 * 22.8 * 0.450
q = 15390 J = 15.4 kJ
This is an endothermic reaction. There is 15.4 kJ of energy required to heat 1.50 kg iron.
