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

Answer : The energy required to heat of 1.50 kg iron is, $1.5\times 10^4J$

Explanation :

Formula used :

$Q=m\times c\times \Delta T$

or,

$Q=m\times c\times (T_2-T_1)$

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:

$Q=1500g\times 0.450J/g^oC\times (15.0-(-7.8))^oC$

$Q=15390J=1.5\times 10^4J$

Therefore, the energy required to heat of 1.50 kg iron is, $1.5\times 10^4J$

Answer 2

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.