50.0 kg rock 12.0℃ heat 0.82 J/g-K 450 kJ heat emit

50.0 kg rock 12.0℃ heat 0.82 J/g-K 450 kJ heat emit

 

 

A) Large beds of rocks are used in

some solar-heated homes to store heat.

Assume that the specific heat of the rocks is 0.82 J/g-K.

Calculate the quantity of heat absorbed by 50.0 kg of rocks

if their temperature increases by 12.0℃.

 

B) Using the information in the above question,

what temperature change would these rocks undergo

if they emitted 450 kJ of heat?

 

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A)

q = C m Δt

( 참고 https://ywpop.tistory.com/2897 )

 

= (0.82 J/g•℃) (50.0 kg) (1000 g / 1 kg) (12.0℃)

= (0.82) (50.0) (1000) (12.0)

= 492000 J

= 492 kJ

 

 

 

[참고] 섭씨온도 1℃와 절대온도 1 K의 간격은 같다.

Δ 1 K = Δ 1 ℃

( 참고 https://ywpop.tistory.com/6662 )

 

 

 

B)

Δt = q / C m

= (450000 J) / [(0.82 J/g•℃) (50.0 kg) (1000 g / 1 kg)]

= (450000) / [(0.82) (50.0) (1000)]

= 10.9756℃

= 11℃