4. A hot pack contains 100 ml salt solution, and it should reach a maximum of 65

Question

4. A hot pack contains 100 ml salt solution, and it should reach a maximum of 65 C. A cold pack contains 100 ml salt solution, and it should reach a minimum of 0 °C In an experiment, 1 gram NH,NO, is dissolved in 30 ml water in a Styrofoam cup. The dissolution causes solution temperature being reduced from 22.5 to 20.4 (a) Can NH NO, be used as hot pack or cold pack? (b) How many grams of NH,NO, are needed for making one pack? 2. If30 mL (-30 g) of warm water (51.0 ) are mixed with 30 mL (-30 g) of cold water (20.2C) in a calorimeter, the equilibrium temperature attained by the system is 34.80. The specific heat of water is 4.181/g Calculate the calorimeter constant for this system.

Explanation / Answer

4. a. As the temperature decreases on mixing NH4NO3 in solution, it can be used to make a cold pack. It is an endothermic reaction.

b. We need cold pack of 100 ml solution for temperature drop of 25.0°C to 0°C. We are taking 25°C as it is room temperature.

1 g dissolved in 30ml gives temperature decrease of 2.1°C.

so for, 0°C we need to dissolve 25/2.1 = 11.9 grams in 30 ml

Now, for to make 100 ml of solution=> 11.9/30 * 100 => 39.7 grams of NH4NO3 will be needed.

2. Calorimeter constant is nothing but the heat capacity of the calorimeter. Let the constant be x J/°C.

Now, calorimeter in the beginning was at the temperature of cold water 20.2°C.

Let us do heat balance.

heat from hot water= heat gained by cold water and calorimeter

30 g * 4.18 J/g°C * (51°C-34.8°C) = 30 g * 4.18 J/g°C * (34.8°C-20.2°C) + x J/°C * (34.8°C-20.2°C)

=> 2031.48 J = 1830.84 J + 14.6x J

=> x = 13.74 J/°C is the heat capacity of the calorimeter.

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