Question

How much heat is required to raise the temperature of 150 gm of a iron ball from 15°C to 25°C ? ( specific heat capacity of 480J!kg0C).

Answer 1

Answer:

Given m=150

g=

1000

150

=0.15kg

Specific heat of iron

C=480Jkg

−1

o

C

−1

)

Δ=(25−20)

o

C=5

o

C

Q=m×C×ΔT

=0.15kg×480Jkg

−1

o

C

−1

×5

o

C

=360J.

Answer 2

ANSWER

_____________________

Given:-

m = 150

$g = \frac{105}{1000} = 0.15kg$

_____________________

Salvation:-

Specific heat of iron,

$C = 480jkg^{ - 1} °C^{ - 1}$

$Δ=(25−20)°C=5°C</p><p></p><p>$

${Q=m×C×ΔT}$

$=0.15kg \: \times 480Jkg ^{ - 1} \:C ^{ - 1} C</p><p> \\ \\ =360J.</p><p></p><p>$