Question

How much energy would be absorbed as heat by 75 g of iron when heated from 295 k to 301 k?

Answer 1

Answer:

q = 202.5 joules

Explanation:

The amount of energy absorbed as one heat 75 grams of iron from 295 k to 301 k can be computed below

q = mcΔt

where

q = energy

m = mass

c = specific heat capacity

Δt = change in temperature

Δt = final temperature - initial temperature

Δt = 301 - 295 = 6 k

c = 0.45  J/g  K

q =  75 g × 0.45 J/g  K × 6 K

g cancels g and k cancels k we are left with the unit in joules

q = 202.5 joules

Answer 2

Answer:

$Q=202.5J$

Explanation:

Hello,

In this case, the increase in the temperature involves the addition of heat that is defined in terms of mass, heat capacity and temperature:

$Q=mCp(T_2-T_1)$

In this case, the heat capacity of iron is 0.450 J/(g*K), thus the heat results:

$Q=75g*0.450\frac{J}{g*K}*(301-295)K\\\\Q=202.5J$

In such a way, since the temperature is increased heat is added, that is why it is positive.

Best regards.