Science [4 Mark Questions]

Total heat = Heat required to convert 2Kg of ice into water at 0°C + Heat required to convert 2Kg of water at 0°C to 2Kg of water at 20°C
Heat = m (hfw) + mc∆T
Here, m(mass of ice) = 2Kg
hfw (specific latent heat of water) = 3,34,000J/Kg
C (specific heat capacity of water) = 4200JKg^{– 1}K^{– 1}
AT (Temperature difference) = 20°C
Therefore, Heat required = (2 × 334000) + (2 × 4200) (20 – 0)
= 668000 + 8400 (20)
= 668000+ 168000
Heat required = 8,36,000 J

To find final temperature: ∆Q = mc
lOOg of water originally at 90°C will loose an amount of heat,
∆Q = mc ∆T
∆Q = 100 × c × (90 – T)
The same amount of heat will be absorbed by 600g of water originally at 20°C to raise its temperature to T.
∆Q = 600 × c × (T – 30)
600C (T – 20°) = 100C (90° – T)
6T – 120° = 90° – T
6T + T = 120° + 90°
7T = 210° ⇒ T = 210/7
T = 30°C