Question

Calculate the energy required to heat of 1.50 kg iron from -7.8 C to 15.0 C . Assume the specific heat capacity of silver under these conditions is .0235 J*g^-1*K^-1 . Be sure your answer has the correct number of significant digits.

Answer 1

There is 15.4 kJ of energy required to heat 1.50 kg iron.

Step-by-step 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.

Answer 2

Answer : The energy required to heat of 1.50 kg iron is,
`1.5* 10^4J`

Explanation :

Formula used :

`Q=m* c* \Delta T`

or,

`Q=m* c* (T_2-T_1)`

where,

Q = heat required = ?

m = mass of iron = 1.50 kg = 1500 g

c = specific heat of iron =
`0.450J/g^oC`

`T_1` = initial temperature =
`-7.8^oC`

`T_2` = final temperature =
`15.0^oC`

Now put all the given value in the above formula, we get:

`Q=1500g* 0.450J/g^oC* (15.0-(-7.8))^oC`

`Q=15390J=1.5* 10^4J`

Therefore, the energy required to heat of 1.50 kg iron is,
`1.5* 10^4J`