200g of ice is heated from temperature T1= - 20°C to T2= + 110°C. a-Calculate the quantity of heat necessary for this transformation?

To calculate the total heat absorbed, we need to consider three stages of heating: heating the ice to 0°C, melting the ice, and heating the water from 0°C to 110°C.

1. Heating the ice from -20°C to 0°C: The specific heat capacity of ice is approximately 2.1 J/g°C. So, the heat (Q1) absorbed can be calculated using the formula Q = mcΔT, where m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. Q1 = 200g * 2.1 J/g°C * 20°C = 8400 J

2. Melting the ice at 0°C to water at 0°C: The heat of fusion of ice is approximately 334 J/g. So, the heat (Q2) absorbed can be calculated using the formula Q = mL, where m is the mass and L is the heat of fusion. Q2 = 200g * 334 J/g = 66800 J

3. Heating the water from 0°C to 110°C: The specific heat capacity of water is approximately 4.18 J/g°C. So, the heat (Q3) absorbed can be calculated using the formula Q = mcΔT, where m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. Q3 = 200g * 4.18 J/g°C * 110°C = 91840 J

The total heat absorbed (Q) is the sum of the heat absorbed in these three stages: Q = Q1 + Q2 + Q3 = 8400 J + 66800 J + 91840 J = 167040 J. So, the quantity of heat necessary for this transformation is 167040 J.