>
#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

Physics

Sort By

ChatGPT

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.

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.