Detail solutions
Asked by swainanirudha1 | 16th Nov, 2018, 04:04: PM
Expert Answer:
Net mass of water to be boiled= 25+1= 26kg
Change in temperature = 100-25 = 75 °C
Specific heat of water= 4186 J/kg °C
Heat required= MCT = 26×75×4186 = 8162700 joule
Answered by Ankit K | 16th Nov, 2018, 10:55: PM
Related Videos
- Is heat a conserved quantity? Give reason in suport of your answer?
- The temperature of 170g of water at 50degree celsius is lowered to 5degree celcius by adding certain amount of ice to it.Find the mass of ice added . (Take ,specific heat capacity of water=4200J/kg0c and specific latent heat of ice =36000J/kg)
- Water falls from a height of 50m. Calculate the rise in the temperature of water when it strikes the bottom. G=10ms-2 C=4200J/Kg°C
Kindly Sign up for a personalised experience
- Ask Study Doubts
- Sample Papers
- Past Year Papers
- Textbook Solutions
Sign Up
Verify mobile number
Enter the OTP sent to your number
Change