A certain amount of heat is required to raise the temperature of x gram of a substance through $t_{1} C$ . The same amount of heat when taken away from y gram of water, it cools through $t_{2} C$ . Therefore, the specific heat of the substance is
$\frac{y t_{1}}{x t_{1}}$
$\frac{x t_{2}}{y t_{1}}$
$\frac{y t_{2}}{x t_{1}}$
$\frac{x t_{1}}{y t_{1}}$

Question

A certain amount of heat is required to raise the temperature of x gram of a substance through $t_{1} C$ . The same amount of heat when taken away from y gram of water, it cools through $t_{2} C$ . Therefore, the specific heat of the substance is
$\frac{y t_{1}}{x t_{1}}$
$\frac{x t_{2}}{y t_{1}}$
$\frac{y t_{2}}{x t_{1}}$
$\frac{x t_{1}}{y t_{1}}$

Toppr Admin · Accepted Answer

For that substance-x394-Q-ms1-x394-T-x-s1-t1-1-Where s1 is specific heat of substanceFor water-x394-Q-m-x2032-s2-x394-T-y-s2-t2-2-Where s2 is specific heat of waterFrom eq -1- and -2- we get-x-s1-t1- - -y-s2-t2-s1s2-yt2xt1Hence- option C is the correct answer and options A is incorrect-

A certain amount of heat is required to raise the temperature of x gram of a substance through t1C. The same amount of heat when taken away from y gram of water, it cools through t2C. Therefore, the specific heat of the substance is yt1xt1xt2yt1yt2xt1xt1yt1