The specific heat capacity of a substance is | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The initial temperature is $T_i = -73^{\circ} C = (-73 + 273) \ K = 200 \ K$.The final temperature is $T_f = 27^{\circ} C = (27 + 273) \ K = 300 \

Vedclass · Accepted Answer

$25000$

Vedclass · Answer

(D) The initial temperature is $T_i = -73^{\circ} C = (-73 + 273) \ K = 200 \ K$.The final temperature is $T_f = 27^{\circ} C = (27 + 273) \ K = 300 \ K$.The heat required $Q$ is given by the integral $Q = \int_{T_i}^{T_f} m C dT$.Given $m = 1 \ kg$ and $C = kT$,we have $Q = \int_{200}^{300} (1) (kT) dT$.$Q = k \int_{200}^{300} T dT = k \left[ \frac{T^2}{2} \right]_{200}^{300}$.$Q = \frac{k}{2} [300^2 - 200^2] = \frac{k}{2} [90000 - 40000] = \frac{k}{2} [50000]$.$Q = 25000 k$.Comparing this with $nk$,we get $n = 25000$.

Similar Questions

Why does the land in a desert get hotter during the day and colder at night quickly?

Why is water used as a coolant in automobile radiators?

Give the $SI$ unit of heat capacity. Write the dimensional formula of heat capacity.

Heat given to a body which raises its temperature by $1\ ^oC$ is

Derive the equation for the molar specific heat capacity of solids.

Vedclass Test Series

Exam Paper Generator

Online Exam Module