Question

6) 100 ml of water is initially at 20°C. 30,000 J of heat is added to the water. What is temperature change for the water?

Answer 1

Δt = 71.67 °C

The temperature change of water is equal to 71.67 °C

Explanation:

Given:

Amount of transferred energy = 30,000 K J

Mass of water = 100 ml

Initial temperature = 20°C

To find the change in temperature of water.

Formula for Heat capacity is given by

Q = m×c×Δt ........................................(1)

where:

Q = Heat capacity of the substance (in J)

m=mass of the substance being heated in grams(g)

c = the specific heat of the substance in J/(g.°C)

Δt = Change in temperature (in °C)

Δt = (Final temperature - Initial temperature) = T(f) - T(i)

Q = 30,000 J

Mass of water = m = 100 ml

1 ml = 1 g ................................................(2)

Therefore m = 100 ml = 100 g

Specific heat of water is c = 4.186 J /g.

Δt = ?

Substituting these in equation (1), we get

Q = m×c×Δt

Rearranging the terms for Δt,

Δt = $\frac{Q}{m\times c}$

Δt = $\frac{30,000}{100\times 4.186} = \frac{30,000}{418.6}= 71.67\°C$

Δt = 71.67 °C