---
title: "Newton's law of cooling: water to 40°C"
subject: "Mathematics I"
chapter: 4
key: "semester_1/mathematics_i:app-newton-cooling-water-40"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/app-newton-cooling-water-40
site: BCT Engineering
---

# Newton's law of cooling: water to 40°C

Mathematics I · Chapter 4: Ordinary Differential Equations and their Applications

- Asked in: 2070 Asar; 2068 Shrawan; 2067 Asar (asked 3 times)
- Video: [Newton's Law of Cooling | First order differential equations | Khan Academy](https://www.youtube.com/watch?v=IICR-w1jYcA) — Khan Academy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/mathematics_i/app-newton-cooling-water-40
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-4

## Question

Newton's law of cooling states that the temperature of an object changes at a rate proportional to the difference of temperature between the object and its surroundings. Supposing water at 100°C cools to 80°C in 10 minutes in a room temperature of 30°C, find the time when the temperature of water will become 40°C.

## Short answer

Using T = 30 + 70e^(−kt) and k = (1/10)ln(7/5), the time to reach 40°C is t = ln7/k ≈ 57.8 minutes.

## Answer

Let T be the temperature of the water at time t (minutes) and let the room temperature be 30°C. Newton's law gives

dT/dt = −k(T − 30), k > 0

#### Solve
dT/(T − 30) = −k dt → ln(T − 30) = −kt + ln C

T − 30 = C e^(−kt)

At t = 0, T = 100: 70 = C. So T = 30 + 70 e^(−kt).

#### Find k
At t = 10, T = 80: 50 = 70 e^(−10k), so e^(−10k) = 5/7 and

k = (1/10) ln(7/5) = 0.03365 per minute.

#### Time to reach 40°C
40 = 30 + 70 e^(−kt) → e^(−kt) = 1/7

kt = ln 7 → t = ln 7/k = 10 ln 7/ln(7/5) = 10(1.9459)/(0.33647)

**t ≈ 57.8 minutes** (about 58 minutes after the start, or 47.8 minutes after the 80°C reading).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which differential equation expresses Newton's law of cooling here (room 30°C)?
   A. dT/dt = k(T + 30)
   B. dT/dt = -k(T - 30)
   C. dT/dt = -k(T - 100)
   D. dT/dt = -kT
   Answer: B. The rate is proportional to the difference between water and room temperature. (Key point: Model)

2. With T(0) = 100, which expression gives T?
   A. 70 + 30 e^(-kt)
   B. 30 + 70 e^(-kt)
   C. 30 - 70 e^(-kt)
   D. 100 e^(-kt)
   Answer: B. T - 30 = C e^(-kt) with C = 100 - 30 = 70. (Key point: General solution)

3. Using T(10) = 80, what equation gives k?
   A. e^(-10k) = 7/5
   B. e^(-10k) = 5/7
   C. e^(-10k) = 4/5
   D. e^(-10k) = 8/10
   Answer: B. 80 = 30 + 70 e^(-10k) gives e^(-10k) = 50/70 = 5/7. (Key point: Finding k)

4. For T = 40°C, the time (from the start) is about:
   A. 20 min
   B. 70 min
   C. 57.8 min
   D. 47.8 min
   Answer: C. e^(-kt) = 1/7, so t = 10 ln7/ln(7/5) = about 57.8 minutes. (Key point: Final time)

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