Advertisements
Advertisements
प्रश्न
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
उत्तर
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
संबंधित प्रश्न
You have a choice of three metals A, B, and C, of specific heat capacities 900 Jkg-1 °C-1, 380 Jkg-1 °C-1 and 460 Jkg-1 °C-1 respectively, to make a calorimeter. Which material will you select? Justify your answer.
What do you mean by the following statement?
The heat capacity of a body is 50 JK-1?
State the effect of enhancement of green house effect.
Solve the following problems:
Equal heat is given to two objects A and B of mass 1 g. Temperature of A increases by 3°C and B by 5°C. Which object has more specific heat? And by what factor?
The ratio of specific heat capacity to molar heat capacity of a body _____________ .
Is it possible to condense the water formed, back to ice by adding ice at 0°C. Explain with reason.
Why are athletes advised to put on extra clothes after competing on event?
The specific heat capacity of water is 1 cal/g °C.
Two metals A and B have specific heat capacities in the ratio 2:3. If they are supplied same amount of heat then
If specific heat capacity of metal A is 0.26 Jg-1 0C-1 then calculate the specific heat capacity of metal B.
