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Solution:

Use Newton's Law of Cooling.  

T = T_s + (T_0 - T_s)*e^(-kt)  

where  

T = temperature at any instant  

T_s = temperature of surroundings  

T_0 = original temperature  

t = elapsed time  

k = constant  

Now, we need to find this constant. We are given that after one hour, the temperature drops to 13° C in a 7°C Environment.  

T = 14, T_0 = 24, T_s = -15, t = 1, k = ?  

T = T_s + (T_0 - T_s)*e^(-kt)  

==> 14 = -15 + (24 - 7)*e^(-k)  

==> 14 = 7 + 17*e^(-k)  

==> 7 = 17*e^(-k)  

==> 7/13 = e^(-k)  

==> -k = ln(7/17)  

==> k = -ln(7/17) ≈ 0.774  

Now,

Let's calculate temperatures!  

T = ?, T_0 = 24, T_s = -15, k = 0.773, t = 3  

T = T_s + (T_0 - T_s)*e^(-kt)  

==> T = -15 + (24 –(-15))*e^[ -(0.774)(2) ]  

==> T = -15 + 39*e^(-1.548)  

==> T ≈ 15.72° C  

This the required answer.