Advertisements
Advertisements
Question
50 g of metal piece at 27°C requires 2400 J of heat energy so as to attain a temperature of 327°C . Calculate the specific heat capacity of the metal.
Advertisements
Solution
m = 50 g or `50/1000` kg
H = 2400 J
Q1 = 27 °C
Q2 = 327 °C
H = mcθ
c = `"H"/("m"theta)`
`= 2400/(50/1000 xx (327 - 27))`
`= 2400/(5/100 xx 300)`
`= 2400/15 = 160` J/Kg - K
APPEARS IN
RELATED QUESTIONS
What do you understand by the following statements:
The heat capacity of the body is 60JK-1.
Explain the term boiling point ?
Explain the meaning of green house effect.
How will climate changes affect the various animal species?
The global warming has resulted:
(a) the increase in yield of crops
(b) the decrease in sea levels
(c) the decrease in human deaths
(d) the increase in sea levels
A certain amount of heat Q will warm 1 g of material X by 3°C and 1 g of material Y by 4°C. Which material has a higher specific heat capacity?
Why are athletes advised to put on extra clothes after competing on event?
How much heat energy is necessary to raise the temperature of 5 kg of water from 20°C to 100°C?
Numerical Problem.
What is the heat in joules required to raise the temperature of 25 grams of water from 0°C to 100°C? What is the heat in Calories? (Specific heat of water = `(4.18"J")/("g"°"C")`
Specific heat capacity of a substance X is 1900 Jkg-1°C-1 means ______.
