1. How much heat must be added to 30.0 g of water to raise the temperature of th

Question

1. How much heat must be added to 30.0 g of water to raise the temperature of the water from 25.5°C to 47.2 °C? 2. What is the sign of the heat change for an exothermic reaction? For an endothermic reaction? 3. A 40.40 g sample of a metal, initial temperature at 100.0 °C, is mixed with 100.0 mL of water, initially at 22.5 °C, in a calorimeter. The water/metal mixture reaches an equilibrium temperature of 24.5 °C. Calculate the specific heat of this metal. The molar heat of solution of a substance is found to be 83.1 kJ/mole. The addition of 0.100 mol of this substance to 1.000 L water initially at 30.0°C results in a temperature increase. Find the final temperature of the solution. Assume that the specific heat of the solution to be equal to that of pure water The heat capacity of the calorimeter is 205 J/oC 4.

Explanation / Answer

Q = mc?T

Q = heat energy (Joules, J), m = mass of a substance (kg)

c = specific heat (units J/kg?K), ? is a symbol meaning "the change in"

?T = change in temperature (Kelvins, K)

Question 1

m = 30 gm ?T = 47.2 -25.5 = 21.7 Deg Cel

Q = 30 x 4.184 x 21.7 = 2723.78 Joules

Hence 2723.78 Joules of heat is required to raise the temperature from 47.2 to 25.5 Deg Cel

Question 2

Exothermic reaction will be notated by negative sign  

Exothermic reaction will be notated by positive sign

Question 3

heat lost by metal = heat gained by water

40.4 x Sp ht x (100 - 24.5) = 100 x 4.184 x (24.5 - 22.5 )

Specific Heat of metal = 0.274 J / g 0C

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