Heat is supplied uniformly at the rate of 100W to 1.0 x 10\(^{-2}\)kg of a liquid for 20 seconds. If the temperature of the liquid rises by 5\(^o\)C, then the specific heat capacity of the liquid is
2.0 x 10\(^2\)Jkg\(^-1\)K\(^-1\)
2.0 x 10\(^2\)Jkg\(^-1\)
4.0 x 10\(^4\)Jkg\(^-1\)K\(^-1\)
4.0 x 10\(^4\)Jkg\(^-1\)
Explanation
Video Explanation
Post your Contribution
Discussions (5)
The steps and all are correct but the options are in kelvin so they should have converted celcius to kelvin however the final answer wouldn't match any option,im saying this so you can have it in the back of your mind.
Alright, let's go step-by-step.
---
**Step 1: Understanding the problem**
- Power supplied \( P = 100 \ \text{W} \)
- Time \( t = 20 \ \text{s} \)
- Mass \( m = 1.0 \times 10^{-2} \ \text{kg} = 0.01 \ \text{kg} \)
- Temperature rise \( \Delta T = 5^\circ \text{C} \)
- We need specific heat capacity \( c \).
---
**Step 2: Total heat supplied**
\[
Q = P \times t = 100 \times 20 = 2000 \ \text{J}
\]
---
**Step 3: Heat equation**
\[
Q = m c \Delta T
\]
\[
2000 = 0.01 \times c \times 5
\]
\[
2000 = 0.05 \times c
\]
\[
c = \frac{2000}{0.05} = 40000 \ \text{J kg}^{-1} \ ^\circ\text{C}^{-1}
\]
---
**Step 4: Final answer**
\[
\boxed{40000}
\]
Specific heat capacity = **40,000 J/(kg·°C)**.
4 X 10⁴J/Kg/C
