Question

1-2.5137g of Mg(NO3)2 is added to 25.0 mL of H2O at 22.5C in a calorimeter. The remperature of the solution rises to 34.8C. calculate the heat of the surroundings.

2- the temperature of 50.0g of hot water in a calorimeter decreases to 45.0C after adding 30.0g of 20C cold water. What was the temperature of the hot water in the calorimeter before adding the cold water?

Answer #1

Q1)

The heat of surroundings = heat of calorimeter

= mass of water x specific heat of water x rise in temperature

= 25 g x 4.184J/g.K x 12.3 K

= 1286.58 J

Q2)

-heat given by hot water = heat absorbed by cold water.

heat given by hot water = mass of hot water sx specific heat x decrease in temperature

= 50 g x 4.184 J/g.K x( 318- t) K

heat absorbed by cold water = 30 g x 4.184 J/g.K x ( 318-293) K

equating

-50 g x 4.184 J/g.K x( 318- t) K = 30 g x 4.184 J/g.K x ( 318-293) K

Thus t = 303 K

= 30C

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Trial 1
Trial 2
Trial 3
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Initial temperature of the cold water in the calorimeter
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21.5
21.5
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50.0g
50.0g
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