Question

3.75 × 10⁵ calories of heat is given out by 5.0 kg of water at 100°C. If the specific heat capacity of water is 1000 cal kg⁻¹°C⁻¹ the temperature of the cold water is:

पर्याय

• 25°C

• 25.5°C

• 24.5°C

• 24°C

Answer 1

25°C

Explanation:

Q = mcΔT

\[\Delta T = \frac{Q}{m\cdot c}\]

\[=\frac{3.75\times 10^{5}}{5.0\times 1000}\]

\[=\frac{3,75,000}{5000}\]

= 75°C

Final temperature = 100°C − 75°C = 25°C