The temperature of a body rises by `44^(@)C` when a certain amount of heat is given to it . The same heat when supplied to `22` g of ice at `-8^(@)C` , raises its temperature by `16^(@)C`. The water eqivalent of the body is `[Given : s_(water) = 1"cal"//g^(@)C` & `L_(t) = 80 "cal" //g, s_("ice") = 0.5"cal"//g^(@)C`]

Options

A. `25`g
B. `50` g
C. `80` g
D. `100` g

Answer

B

Explanation

Supplied heat = `(22) (0.5) (8) + (22)(80) + (22)(1)(16) = 88 + 1760 +352 = 2200"cal"`
Heat capacity of the body `=(2200"cal")/(44^(@)C) = 50 "cal"//^(@)C`
Water equavalent of the body =` ("Heat capacity of the body")/("specific heat capacity of water") = (50"cal"//^(@)C)/(1 "cal"//g^(@)C) = 50"g"`