The normal cooling equation (Newton) has the form:
Cooling body temperature, K(t)=tA+(tH-tA)e-kt, or K(t)=(tH-tA)e-kt+tA, where k is a constant, TH is the initial temperature of the cooling body (in this case, a sandwich) and TA is the ambient or room temperature. After a long period the body cools to the ambient temperature. The given equation should resemble this standard equation. tH-tA can be replaced by a single quantity which I think is 60 in the given equation. This would mean the initial temperature difference between the internal temperature of the sandwich and the room temperature is 60°C.
e-kt can be replaced by bt, that is, a number raised to power x. -kt=ln(bt)=tln(b), making k=-ln(b). I think that b=0.5 in the given equation. So now we have an explanation for the term 60(0.5)t. The remaining expression in the given equation is 10+2. This suggests that the ambient temperature TA is 2°C. That leaves the value 10. I suggest that the given equation is meant to read:
K(t)=60(0.5)t/10+2, which is an acceptable variation of the standard cooling equation. I think this is the cooling equation. Bearing that in mind we can look at the questions.
(a) When t=0, K(0)=62°C.
(b) K(20)=60(0.5)2+2=17°C.
(c) K(t)=32, 60(0.5)t/10=30, 0.5t/10=0.5, so t/10=1, t=10 minutes.
(d) Absolute zero is -273°C approx, but we're not going to need this. The cooling equation indicates that after an indefinitely large time, K=2°C. So the domain is [0,∞) or 0≤t<∞. The range for this domain is K(0)=62°C down to nearly 2°C, that is (2,62] °C or 2<K≤62 (°C).
