How much steam at 100 degree centigrade will just melt 3200g of rice at -10 degree centigrade ? ( S.H.C of rice = 0.5 cal/g digree centigrade; Let heat of vapour of steam =540cal/gg).
To find out how much steam at 100 degrees Celsius is needed to melt 3200 grams of rice at -10 degrees Celsius, you need to calculate the amount of heat required to raise the temperature of the rice from -10 degrees Celsius to 0 degrees Celsius and then the amount of heat to convert the rice from solid to liquid.
1. Calculate the heat required to raise the temperature of the rice from -10 degrees Celsius to 0 degrees Celsius:
Heat = mass × specific heat capacity × change in temperature
Heat = 3200 g × 0.5 cal/g°C × (0°C - (-10°C))
Heat = 3200 g × 0.5 cal/g°C × 10°C
Heat = 16000 cal
2. Calculate the heat required to convert the solid rice at 0 degrees Celsius to liquid rice at 0 degrees Celsius:
Heat = mass × heat of vaporization
Heat = 3200 g × 540 cal/g
Heat = 1728000 cal
3. Add the two amounts of heat together to get the total heat required:
Total heat = 16000 cal + 1728000 cal
Total heat = 1744000 cal
Now, the heat required to raise the temperature of the water to 100 degrees Celsius is negligible compared to the heat of vaporization, so we can ignore that part.
4. Calculate the amount of steam needed using the heat of vaporization:
Amount of steam = Total heat / heat of vaporization
Amount of steam = 1744000 cal / 540 cal/g
Amount of steam = 3233.33 g
Therefore, you would need approximately 3233.33 grams (or 3.233 kg) of steam at 100 degrees Celsius to just melt 3200 grams of rice at -10 degrees Celsius.
Melt rice??
You need latent heat of fusion to do this one. And specific heat of rice.
PS. Are you trying to learn all of Physics in one night?