8. a. Write the two mathematical relationships for the heat Q in two cases: with

Question

8. a. Write the two mathematical relationships for the heat Q in two cases: with a change temperature (delta t= delta T not equal to 0) and without this change (delta t= delta T = 0). Comment them. Use these formula in the following problem: A mass m=1kg of crushed ice, initially at -10oC, was placed on the kitchen table and melted. How much energy is needed to melt
the ice? Where does the ice get this energy? lf the same amount of energy calculated in the first part was used to heat water at O°C to 50°C, how much water. (its mass m.) could be heated? (1 cal=4186 J; c(specific heat of ice)=.5cal/oC, g=2093J/oC kg; latent of heat melting=79.8 cal/g=334kJ/kg; c(specific heat of water)=1cal/0C g=4186J/oC kg)

8. b. Calculate the increase in temperature delta t=delta T of water which falls from a height h=100 m, assuming that all the energy due to fall is converted into heat and is retained by water. [g = 9.8m/s2, 1kcal=4186 J, c(specific beat of water=1cal/oC g= 4186 J/oC kg].

8. c. What are the main mechanisms regulating the temperature of the human body? Mention and comment them.

Explanation / Answer

For delta t = delta T = 0, the heat exchange = M*L where L is Latent heat because temperature is not changing here and the heat is given to change the state of matter. In case of delta T not equal to zero, that means the temperature is being change the formula for heat transfer is given by Q = ms delta T where s is specific heat.

1kg ice at -10 deg C is melted.

We know that ice is melted in water only at 0 deg C hence the energy required will be,

Q = m s delta T + m*L = 1000*0.5*10 + 1000*79.8 = 84800 cal.

The ice gets this energy from the surroudings.

The amount of water is m = Q/s*delta T = 84800/(1*50) = 1696g = 1.696 kg.

8. b. Energy given = mgh = energy taken = m s delta T

delta T = gh/s = 9.8*100/(4186) = 0.23 deg C.

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