The graph shows the temperature versus time relationship for 2kg of particular substance. The substance is in the solid state at a temperature of 20 degrees Celsius
b) If 200 joules of heat is added during the time interval B(20 degrees Celsius) to C(80 degrees Celsius), what is the specific heat of the substance
heat=m*c*dT
c= 200j/*2kg60C)= 200/(120) J/gC
ooops, J/KgC
Is the answer 1.67
To find the specific heat of the substance, we can use the formula:
Q = mcΔT
where:
Q is the amount of heat added or removed (in joules)
m is the mass of the substance (in kg)
c is the specific heat capacity of the substance (in joules per kilogram per degree Celsius)
ΔT is the change in temperature (in degrees Celsius)
In this case, we know the following information:
Q = 200 J
m = 2 kg
ΔT = 80 - 20 = 60 degrees Celsius
Now we can rearrange the formula to solve for c:
c = Q / (m * ΔT)
Substituting the known values:
c = 200 J / (2 kg * 60 degrees Celsius)
c = 200 J / 120 kg°C
Dividing both the numerator and denominator by 40:
c = 5 J / 10 kg°C
c = 0.5 J / kg°C
Therefore, the specific heat of the substance is 0.5 joules per kilogram per degree Celsius.