104 g of water at 30 degree celsius is taken in a calorimeter made of copper of mass 42 gram when a certain mass of ice at zero degree Celsius is added to it a final study temperature of the mixture of the devices melted was found to be 10 degree Celsius find the mass of ice added specific heat capacity of water 4.2 joule per gram per Celsius specific latent heat of fusion of ice 336 joule per gram specific heat capacity of copper 0.4 joule per gram per degree Celsius
Answers
From method of mixtures
Heat lost by water and copper vessel is equal to the heat gained by ice
Applying
[104×4.2×(30-10)]+[42×0.4×(30-10)]=
(m×336)+(m×4.2×(10-0))
= 8736+336=m(378)
=9072. =m(378)
Therefore m=9072÷378
=24g
Here, we can solve this problem by applying principle of thermal equilibrium or calorimetry .
According to principle of thermal equilibrium , when a higher temperature object comes in contact with a lower temperature object, it will transfer heat to the lower temperature object. Both the objects will achieve same temperature, until they maintain a constant temperature.
Here, ice added will first be melted so , we will take latent heat of fusion of ice .Than, the water formed will gain heat .
Heat lost or Heat gained = mass of substance * specific heat of substance * change in temperature .
Here, heat gained by ice = heat lost by water + heat lost by calorimeter
=> (336 * m) + (m * 4.2 * 10) = 104 * 4.2 * (30 - 10) + [42 * 0.4 * (30 - 10)]
=> 378m = 8736 + 336
=> m = 24 gm
Thus, the mass of ice added = 24 gm