Advertisements
Advertisements
Question
Calculate the mass of ice required to lower the temperature of 300 g of water 40°C to water at 0°C.
(Specific latent heat of ice = 336 J/g, the Specific heat capacity of water = 4.2J/g°C)
Solution
Let m be the mass of the ice to be added.
Heat energy required to melt to lower the temperature is = mL = m x 336
Heat energy imparted by the water in fall of its temperature from 40°C to
0°C = mass of the water x specific heat capacity x fall in temperature
= 300 x 4.2 x 40°C
If there is no loss of heat,
m x 336 J/g = 300 g x 4.2 J/g°C x 40°C
`:. m = (300 xx 4.2 xx 40)/336`
∴ m = 150g
APPEARS IN
RELATED QUESTIONS
Calculate the mass of ice needed to cool 150 g of water contained in a calorimeter of mass 50 g at 32 °C such that the final temperature is 5 °C. Specific heat capacity of calorimeter = 0.4 J g-1 °C-1, Specific heat capacity of water = 4.2 J g-1°C-1, latent heat capacity of ice = 330 J g-1.
State S.I. unit of specific heat capacity.
What do you mean by the following statement?
The heat capacity of a body is 50 JK-1?
Name three fossil fuels that emit carbon dioxide into the atmosphere ?
State the condition for the flow of heat energy from one body to another.
Derive an expression for finding out the specific heat capacity of a body (solid) from the readings of an experiment given below:
(i) Mass of empty calorimeter (with stirrer) = m1 gm
(ii) Mass of the metal piece = M gm
(iii) Mass of colorimeter and water = m2 gm
(iv) Initial temperature and water = t1°C
(v) Temperature of hot solid (metal piece) = t2 °C
(vi) Final temperature of the mixture = t°C
(vii) Specific heat of calorimeter = 0.4 J gm / °C
Answer the following question.
Why do we generally consider two specific heats of a gas?
Which of the following substances (A, B and C) has the highest specific heat?
Prove the Mayer's relation `C_p - C _v = R/J`
