Little ice age and sunspots

By the formula $P = kT^4$ , where $k$ are product of constants that includes emissivity, area, and Stefan-Boltzmann constant, this proportional relation can be used to solve the problem.

Hence, by $\frac{P_{today}}{P_{LittleIceAge}} = \frac{T_{today}^4}{T_{LittleIceAge}^4}$ , plugging in

$P_{today} = 1361.8 W, P_{LittleIceAge} = 1360.15, T_{today} = 287 K$ , the value of $T_{LittleIceAge} = 286.91 K$

Hence, by $T_{today} - T_{LittleIceAge}$

$= 287 - 286.91$

$= \boxed{0.87} K$

The Stefan-Boltzmann law (which I shan't derive since it's well-available on the Wikipedia page ) dictates that, for some energy radiated per area, $j^*$ and some thermodynamic temperature $T$ : $j^*=\sigma T^4$

is true for the output for the temperature of a blackbody.

Since we are assuming equilibrium, then the converse also holds, but note that the Earth is not completely under sunlight, hence our constant, $\sigma$ , doesn't. So, the most restrictive case we have is: $j^*\propto T^4$

Since we are given the temperature and irradiance for one case, we can compute the proportionality constant $l$ , by: $j_0^*=lT_0^4$

Hence: $\frac{j_0^*}{T_0^4}=l=\frac{1361.8}{287^4}$

So, for our final case gives us: $\sqrt[4]{\frac{j^*}{l}}=T=\sqrt[4]{\frac{287^4}{1361.8}\cdot 1360.15}\approx286.91303\text{ K}$

Which, finally yields: $T_0-T\approx 287-286.91303=0.08697\text{ K}$

Our answer.

We know from Stefan's law that P = aAT^{4} where P = Power a = stefan's constant T= temperature

In this question we are given P/A ratios. SO taking ratio of these ratios- 1360.15/1361.8 = (T/287)^{4} We get T = 286.013 SO T1 - T2 = 0.08697

For a black body ratio of solar irradiance(E) and fourth power of temperature(T^4) remains constant, i.e.,

E/(T^4) = constant(say k)

Let E1 and T1 be irradiance and temperature of Eath Little Ice Period and E2 and T2 be irradiance and temperature of Earth today.

(E1)/((T1)^4) = (E2)/((T2)^4)

((T1)^4) = (1360.15/1360.8) * (287^4)

T1 = 286.913 K

T2 - T1 = 287 - 286.913 = 0.087 (Ans)