Question

1) How many kJ are absorbed when 45.2 g of water at 31.3 oC is heated to 76.9 oC? 2) Calculate the total heat released in kcal when 72.1 g water at 25.2 oC is cooled to 0 oC and freezes. 3) How many kilojoules are required to heat 55,500 mg of gold with specific heat = 0.129 J/g oC is heated from 24.6 oC to 123.4 oC? 4) Calculate the heat needed in kcal to change 45.6 g of water at 100 oC to change into steam.

Answer 1

Answer:

1. Q = 8.66 KJ

2. Q = 7.58 Kcal

3. Q = 0.71 KJ

4. Q = 24.31 Kcal

Explanation:

1. The quantity of heat absorbed can be determined by:

       Q = mcΔθ

where: Q is the quantity of heat absorbed or released, m is the mass of the substance, c is the specific heat capacity of water = 4.2 j/g$^{0}C$ and Δθ is the change in temperature.

           = 45.2 × 4.2  × (76.9 - 31.3)

           = 8656.704

∴       Q = 8.66 KJ

The quantity of heat absorbed is 8.66 KJ.

2. Q = mcΔθ + mL

Where L is the latent heat of fusion of ice = 334 J.

       = m(cΔθ + L)

       = 72.1(4.2 × 25.2 + 334)

   Q = 31712.464 J

       = 7579.466 calories

The total heat released is 7.58 Kcal.

3. Q = mcΔθ

      = 55.5 × 0.129 × (123.4 - 24.6)

     = 707.3586

The quantity of heat required to increase the temperature of gold is 0.71 KJ.

4. Q = mL

Where: L is the specific latent heat of vaporization = 533 calories.

     Q  = 45.6 × 533

      = 24304.8

The quantity of heat required to change water to steam is 24.31 Kcal.