Question

# 1-2.5137g of Mg(NO3)2 is added to 25.0 mL of H2O at 22.5C in a calorimeter. The...

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?

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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