1) A 100.0-g sample of water at 27.0 o C is poured into a 76.2-g sample of water

Question

1) A 100.0-g sample of water at 27.0oC is poured into a 76.2-g sample of water at 89.0oC. What will be the final temperature of the water?

2) A calorimeter contains 240 g of water at 21.5°C. A block of metal with a mass of 94 g is heated to 97.3°C and then placed in the water in the calorimeter. After sufficient time, the temperature of the water is measured and found to be 26.9°C. Calculate the specific heat capacity per gram of metal. Assume no heat is lost to the calorimeter or the surroundings.

Explanation / Answer

1)Heat lost by hot water=Heat gained by cold water

Let the final temperature be T

So

(76.2)(4.186)(89-T)=(100)(4.186)(T-27)

6781.8-76.2T=100T-2700

T=53.81 deg C

2)Heat lost by the metal =Heat gained by water

Let the specific heat do the metal be S

So

(94)(S)(97.3-26.9)=(240)(4.186)(26.9-21.5)

Silving we get

S=0.8197 J/gm-C or 819.7 J/Kg-C

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