How many grams of water can be healed from 28 degree C to 73 degree C with 500.0

Question

How many grams of water can be healed from 28 degree C to 73 degree C with 500.0 calories of heat? When 300.0 calories of heat are added to 32-5 g of copper (specific heal of 0.0920 cal/g degree C), what is the temperature change? A sample of metal has a mam of 1.75 kg and a specific heat of 0.115 cal/g degree C. What is the temperature change of the metal after 625 calories of heat are added? If 120.0 g of copper at 100.0 degree C is added to 28.6 grams of water at 26.5 degree C. what is the final equilibrium temperature?

Explanation / Answer

(1)

Specific heat of water = 1 cal/g oC

We know,

mC(T2-T1) = heat absorbed

m x 1x(73-28) = 500

m = 11.11 grams

(2)

We know,

mC(T2-T1) = heat absorbed

32.5 g x0.092x(T2-T1) = 300

(T2-T1) = 92.64 oC

(3)

We know,

mC(T2-T1) = heat absorbed

1750 g x0.115 x(T2-T1) = 625

    (T2-T1) = 3.11 oC

(4)

Let the fibal temperature = T2

Heat lost by copper = heat gained by water

120.0 g x 0.095 x (100-T2) === 28.6 g x 1 x (T2-26.5)

T2 =47.44 oC

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