4200 J of work is required for:

CONTENT:

Specific heat:

• It is defined as the amount of heat required to increase the temperature by 1°K for 1 kg of material.
• So the heat required to increase the temperature by Δt is given as,

⇒ Q = mC∆t     -----(1)

Where Q(J) = quantity of heat absorbed by a body, m(kg) = mass of the body, ∆t(°K) = Rise in temperature, and C = Specific heat

• S.I unit of specific heat is t is J kg-1 K-1.

CALCULATION:

We know that the specific heat of water is 4200 J/kg-K.

For option 1: (m = 10 gm = 10^-2 kg, Δt = 10° C)

⇒ Q = mCΔt

⇒ Q = 10^-2 × 4200 × 10

⇒ Q = 420 J

For option 2: (m = 100 gm = 10-1 kg, Δt = 10° C)

⇒ Q = mCΔt

⇒ Q = 10-1 × 4200 × 10

⇒ Q = 4200 J

For option 3: (m = 1 kg, Δt = 10° C)

⇒ Q = mCΔt

⇒ Q = 1 × 4200 × 10

⇒ Q = 42000 J

For option 4: (m = 10 kg, Δt = 10° C)

⇒ Q = mCΔt

⇒ Q = 10 × 4200 × 10

⇒ Q = 420000 J

• Since the heat required in option 2 is 4200 J so we can say that 4200 J of work is required for increasing the temperature of 100 gm of water through 10° C. Hence option 2 is correct.