$A$ black body radiates maximum energy at wavelength $\lambda$ and its emissive power is $E$. Now, due to a change in the temperature of that body, it radiates maximum energy at wavelength $\frac{2 \lambda}{3}$. At that temperature, the emissive power is:

If $120 \ J$ of thermal energy is incident on an area of $3 \ m^2$,the amount of heat transmitted is $12 \ J$,and the coefficient of absorption is $0.6$,then the amount of heat reflected is: (in $J$)

$A$ hot black body emits energy at the rate of $16 \ J \ m^{-2} \ s^{-1}$ and its most intense radiation corresponds to $20,000 \ \mathring{A}$. When the temperature of this body is further increased and its most intense radiation corresponds to $10,000 \ \mathring{A}$,then the energy radiated in $J \ m^{-2} \ s^{-1}$ will be

What are the modes of heat transfer due to a temperature difference?

Fill in the blanks:
$(a)$ $0.49 \frac{\text{cal}}{\text{cm} \cdot \text{K} \cdot \text{s}} = \dots \frac{\text{J}}{\text{m} \cdot \text{K} \cdot \text{s}}$
$(b)$ If the rate of emission of heat of a substance is less than its rate of absorption,then its temperature $\dots$.
$(c)$ The rate of emission of heat of a substance is directly proportional to $\dots$ of temperature of it and surroundings.

$A$ heated body emits radiation which has maximum intensity at frequency $f_m$. If the temperature of the body is doubled,then: