The specific heat of a certain type of metal is 0.128 J/(g·°C). What is the final temperature if 305 J of heat is added to 83.0 g of this metal initially at 20.0 °C?

Answer 1

We are aware of the heat gain.

#DeltaH=mxxsxx(t_2-t_1)#

where

#DeltaH=305J# #m->"mass"=83g#
#s->"sp heat"=0.128" J/(g^@C)#
#t_1->"initial temperature"=20^@C#
#t_2->"final temperature"=?#

Thus

#DeltaH=mxxsxx(t_2-t_1)#
#=>305J=83gxx0.128J/(g^@C)xx(t_2-20)^@C#
#=>t_2-20=305/(83xx0.128)#
#=>t_2=48.7^@C#
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Answer 2

The final temperature of the metal is 68.0 °C.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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