Question

Calculate the amount of heat gained when one 250 gram bottle is heated from 25oC to 30oC. The specific heat of water is 4.18 J/goC

Answer 1

Answer:

5230J

Explanation:

Mass (m) = 250g

Initial temperature (T1) = 25°C

Final temperature (T2) = 30°C

Specific heat capacity (c) = 4.184J/g°C

Heat energy (Q) = ?

Heat energy (Q) = Mc∇T

Q = heat energy

M = mass of the substance

C = specific heat capacity

∇T = change in temperature = T2 - T1

Q = 250 × 4.184 × (30 - 25)

Q = 1046 ×5

Q = 5230J

The heat energy required to raise the temperature of 250g of water from 25°C to 30°C is 5230J

Answer 2

Answer:

The amount of heat gained is 5225 J

Explanation:

The specific heat of a substance is the quantity or amount of heat required to change the temperature of a unit mass of the substance by 1$^{0}C$.

Thus,

         Q = mcΔθ

where: Q is the quantity of heat required, m is the mass of the substance, c is the specific heat capacity of the substance and Δθ is the change in temperature of the substance.

From the question, given that: m = 250 g, initial temperature = 25$^{0}C$, final temperature = 30$^{0}C$ and specific heat of water = 4.18 J/g$^{0}C$.

So that; the quantity of heat, Q, required is;

         Q = 250 × 4.18 × (30 - 25)

             = 250 × 4.18 × 5

            = 5225 Joules

⇒     Q = 5225 J

Therefore, the amount of heat gained is 5225 J.