Question

calculate the heat required to convert 20 g of ice at -5 degree celcius to water at 50 degree celcius

Answer 1

Answer:

m=20g

temp=50-5=45

c specific heat capcity of water= 4.2 j per kg k

Q (heat energy)=mxcxt=4.2x20x45=3780J

Explanation:

Answer 2

The heat required is 2650 cal.

The amount of ice (m) is = 20 g

The initial temperature of ice (T₁) = -5°C

The final temperature of water (T₂) = 50°C

We know,

The latent heat of ice (L) = 80 cal/g

The specific heat of ice (s) = 0.5 cal/g °C

The specific heat of water (s) = 1 cal/g °C

Heat required, H = H₁ + H₂ + H₃

H₁ = Heat required to convert ice of -5°C to 0°C

∴ H₁ = msΔT

⇒  H₁ = 20 × 0.5 × 5 cal

⇒ H₁ = 50 cal

H₂ = Heat required to convert 0°C ice to 0°C water.

H₂ = mL

⇒ H₂ = 20 × 80 cal

⇒ H₂ = 1600 cal

H₃ = Heat required to convert 0°C water to 50°C water.

∴ H₃ = msΔT

⇒ H₃ = 20 × 1 × 50 cal

⇒ H₃ = 1000 cal

So, the total heat required will be = (50 + 1600 +1000) cal

                                                       = 2650 cal