Question

The specific heat of a substance varies with temperature as t°C as (0.2t + 3)calºc-1 . Heat required to raise the temperature body made of this substance from 0°C 10°C is (A) 2000 cal th (B) 4000 cal (C) 5000cal (D) 3000 cal​

Answer 1

Answer:

40 cal

Explanation:

From the Heat, specific heat equation, we know that

$\frac{dQ}{dT} = C\\\\dQ = C dT\\\\Q = \int\limits^{10}_{0} {(0.2T + 3)} \, dT\\\\Q = [0.2\frac{T^2}{2} + 3T]^{10}_{0}$

Q = 0.1T² + 3T

Q = 0.1 × 100 + 30

Q = 40 cal

(To get 4000 cal, temperature should be raised to 100°C not 10)