> #heat-transfer , #thermodynamics , #physics > Calculate the specific heat capacity of the materials for which the heat supply mass and temperature rise are .i 2.52*10⁴j ,20kii 8*10³j 500g from 20c⁰ to 60c⁰iii 9.45 *10³j 150g from 15c⁰ to 45c⁰ ??
Mar 23

# Calculate the specific heat capacity of the materials for which the heat supply mass and temperature rise are .i 2.52*10⁴j ,20kii 8*10³j 500g from 20c⁰ to 60c⁰iii 9.45 *10³j 150g from 15c⁰ to 45c⁰ ??

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Mar 23
To calculate the specific heat capacity of a material, we can use the formula:

Q = mcΔT

Where:
Q = heat supplied (in Joules)
m = mass of the material (in grams)
c = specific heat capacity of the material (in J/g°C)
ΔT = temperature change (in °C)

i) For the first scenario:
Q = 2.52*10⁴ J
m = unknown
ΔT = 20°C

ii) For the second scenario:
Q = 8*10³ J
m = 500g
ΔT = 40°C

iii) For the third scenario:
Q = 9.45*10³ J
m = 150g
ΔT = 30°C

Now, we can calculate the specific heat capacity for each scenario:

i) 2.52*10⁴ = m * c * 20
c = (2.52*10⁴) / (m * 20)

ii) 8*10³ = 500 * c * 40
c = (8*10³) / (500 * 40)

iii) 9.45*10³ = 150 * c * 30
c = (9.45*10³) / (150 * 30)

Now, we can calculate the specific heat capacity for each scenario using the given information.
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