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 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.
