Question

A metal sheet of mass 1.2 kg is heated to increase its temperature from 400C to 900C. The

amount of heat given is 80 J then find its specific heat capacity.​

Answer 1

answer : 0.133 J/Kg/k

explanation :

• mass of of metal sheet is 1.2 kg

• given heat energy is 80J

• heated from 400 C to 900C

• changing temperature {∆t}

∆t = 900-400 = 500 C

• specific heat = Q/m∆t

= 80/1.2×500

= 8/60

= 0.13333..j/ kg/k

Answer 2

Answer:

Answer:The answer of your question is 0.133 J/kg/K or 0.133 Jkg-K-

Answer:The answer of your question is 0.133 J/kg/K or 0.133 Jkg-K-Explanation:

Answer:The answer of your question is 0.133 J/kg/K or 0.133 Jkg-K-Explanation:Given:Mass of sheet=1.2 kg

Answer:The answer of your question is 0.133 J/kg/K or 0.133 Jkg-K-Explanation:Given:Mass of sheet=1.2 kgChange in temperature (∆t)=900°C-400°C=500°C=500 K ( ∆t °C=∆t K)

Amount of heat energy(Q)=80 J

Sol.)We know that,Specific heat capacity(c)= Q/m∆t

Sol.)We know that,Specific heat capacity(c)= Q/m∆t=. 80/(1.2*500)

Sol.)We know that,Specific heat capacity(c)= Q/m∆t=. 80/(1.2*500)=. 8/(1.2*50)

Sol.)We know that,Specific heat capacity(c)= Q/m∆t=. 80/(1.2*500)=. 8/(1.2*50)=. 2/(0.3*50)

Sol.)We know that,Specific heat capacity(c)= Q/m∆t=. 80/(1.2*500)=. 8/(1.2*50)=. 2/(0.3*50)=. 1/(0.3*25)

Sol.)We know that,Specific heat capacity(c)= Q/m∆t=. 80/(1.2*500)=. 8/(1.2*50)=. 2/(0.3*50)=. 1/(0.3*25)=. 1/7.5

Sol.)We know that,Specific heat capacity(c)= Q/m∆t=. 80/(1.2*500)=. 8/(1.2*50)=. 2/(0.3*50)=. 1/(0.3*25)=. 1/7.5=. 0.133 J/kg/K

Hence,specific heat capacity of the metal sheet= 0.133 J/kg/K