After how long will the temperature of a body heated to 100, decrease to 25 C

After how long will the temperature of a body heated to 100, decrease to 25 C, if the room temperature is 20 C and in the first 10 minutes the body has cooled to 60 C?

1. Let:

T1 = 100 ° C – initial body temperature;
T2 = 60 ° C – body temperature after 10 minutes;
T3 = 25 ° C – final body temperature;
T0 = ​​20 ° C – room temperature.
2. The rate of body cooling is proportional to the difference in body and ambient temperatures:

dT / dt = k (T – T0), hence:
dt = dT / (k (T – T0));
dt = -d (T – T0) / (k (T – T0));
t (T) = -∫d (T – T0) / (k (T – T0)) = -1 / k * ∫d (T – T0) / (T – T0) = -1 / k * ln (T – T0) + C;
{t (T1) = 0;
{t (T2) = 10;
{-1 / k * ln (T1 – T0) + C = 0;
{-1 / k * ln (T2 – T0) + C = 10;
{-1 / k * ln (100 – 20) + C = 0;
{-1 / k * ln (60 – 20) + C = 10;
{-1 / k * ln80 + C = 0;
{-1 / k * ln40 + C = 10;
{C = 1 / k * ln80;
{-1 / k * (ln40 – ln80) = 10;
{C = 1 / k * ln80;
{1 / k * (ln80 – ln40) = 10;
{C = 1 / k * ln80;
{1 / k * ln2 = 10;
{C = 10 / ln2 * ln80;
{1 / k = 10 / ln2;
{C = 10ln80 / ln2;
{1 / k = 10 / ln2;
t (T) = -1 / k * ln (T – T0) + С;
t (T) = -10 / ln2 * ln (T – T0) + 10ln80 / ln2;
t (T) = 10ln80 / ln2 – 10 / ln2 * ln (T – T0);
t (T) = 10 / ln2 (ln80 – ln (T – 20));
t (T) = 10 / ln2 * ln (80 / (T – 20));
t (T) = 10log2 (80 / (T – 20)).
3. Check the values ​​of t (100), t (60) and calculate t (T3):

t (100) = 10log2 (80 / (100 – 20)) = 10log2 (1) = 0;
t (60) = 10log2 (80 / (60 – 20)) = 10log2 (80/40) = 10log2 (2) = 10;
t (25) = 10log2 (80 / (25 – 20)) = 10log2 (80/5) = 10log2 (16) = 10 * 4 = 40 (min.).
Answer: after 40 minutes.