Question

The specific heat capacity of lead is 0.13. how much heat in j is required to raise the temperature of 15 g of lead from 22degrees c to 37 degrees c

Answer 1

Answer:

Specific heat (C)=0.13 J/g-K Mass(m)=15g Initial temperature(T1)=22C=295K : final temperature(T2)=37C=310K Chage in.

Answer 2

Final answer:

The amount of heat is required to raise the temperature of 15 g of lead from 22 °C to 37 °C, if specific heat capacity of lead is $0.13 Jg^{-1}K^{-1}$  = 29.25 J

Given that: We are given the specific heat capacity of lead is $0.13 Jg^{-1}K^{-1}$.

To find: We have to find how much heat in J is required to raise the temperature of 15 g of lead from 22 °C to 37 °C.

Explanation:

• Specific heat capacity of substance is the quantity of heat required to raise the temperature of that substance by 1 g.

Specific heat capacity, $c = \frac{q}{m\triangle T}$

• $q = c m\triangle T$

Here,

Specific heat capacity of lead $c = 0.13 Jg^{-1}K^{-1}$

m = 15 g

$\triangle T=T_{2}-T_{1}$

      $T_{2} = 37 C = 273+37 = 310 K\\\\T_{1} = 22 C = 273+22 = 295 K$

$\triangle T = (310K-295K) = 15K$

q = Heat energy = ?

• Substitute the value of c, m, ΔT.

$q = 0.13 \frac{J}{g.K}* 15g*15K$

  $= 29.25$ Joule

• q = 29.25 J

• Hence, the amount of heat is required to raise the temperature of 15 g of lead from 22 °C to 37 °C, if specific heat capacity of lead is $0.13 Jg^{-1}K^{-1}$  = 29.25 J

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