Advertisements
Advertisements
Question
650 J of heat is required to raise the temp. of 0.25 kg of lead from 15°C to 35°C. Calculate the Sp. heat capacity of lead.
Advertisements
Solution
Q = 650 J
m = 0.25 kg
ΔT = (35 - 15) = 20°C
Q = m x C x T
C = `"Q"/("m" xx Δ"T") = 650/(0.25 xx 20) = 130` J/kg°C
APPEARS IN
RELATED QUESTIONS
During the phase change does the average kinetic energy of the molecules of the substance increase?
Explain, why does a wise farmer water his fields, if forecast is forst?
The heat capacity of the vessel of mass 100 kg is 8000 J/°K. Find its specific heat capacity.
What is specific heat capacity?
The ratio of the specific heats `c_"p"/c_"v"=gamma` in terms of degrees of freedom 'n' is given by ______.
Two metals A and B have specific heat capacities in the ratio 2 : 3. If they are supplied the same amount of heat then
Which metal piece will show a greater rise in temperature given their masses is the same?
To study energy exchange between hot and cold objects, the system of both objects is isolated from the environment by keeping them inside ______.
Two blocks P and Q of different metals having their mass in the ratio 2 : 1 are given same amount of heat. Their temperature rises by same amount. Compare their specific heat capacities.
