Question

I forgot to attach the photo before.... please help

Answer 1

Part (i)

I'll use H in place of T to represent the heat of the object. That way there isn't a clash of variables lowercase t vs uppercase T.

The equation we're working with is updated to:

H(t) = 22 + a*2^(bt)

Plugging in t = 0 as the initial time value should lead to the temperature being H = 86 degrees Celsius.

So,

H(t) = 22 + a*2^(bt)

86 = 22 + a*2^(b*0)

86 = 22 + a*2^0

86 = 22 + a*1

86 = 22 + a

a+22 = 86

a = 86-22

a = 64

Answer:  64

=====================================================

Part (ii)

We'll use the value of 'a' we found earlier. Plus we'll use the fact that H = 28 when t = 0.5 (since 30 min = 30/60 = 0.5 hr).

H(t) = 22 + a*2^(bt)

28 = 22 + 64*2^(b*0.5)

28-22 = 64*2^(0.5b)

64*2^(0.5b) = 6

2^6*2^(0.5b) = 6

2^(6+0.5b) = 6

log(  2^(6+0.5b)   ) = log(6)

(6+0.5b)*log(2) = log(6)

6+0.5b = log(6)/log(2)

6+0.5b = 2.5849625

0.5b = 2.5849625-6

0.5b = -3.4150375

b = -3.4150375/(0.5)

b = -6.830075

Answer: Approximately   -6.830075