Can somebody help me out solving this question step by step

Can somebody help me out solving this question step by step please? I want to study not just the answer please.

A cup of coffee is heated in a microwave oven. If the mass of coffee (modeled as liquid water)
is 0.2 kg and the rate of heat transfer is 0.1 kW, (a) determine the rate of change of internal
energy (u). (b) Assuming the density of coffee is 1000 kg/m3  and the specific internal energy in
kJ/kg is related to temperature through u = 4.2T, where T is in Kelvin, determine how long it
takes for the temperature of the coffee to increase by 20°C.

Solution

Physics involved: The oven heating power is 0.1 KW. Meaning the oven radiates 0.1 KJ of heat per second. and this amount of heat is supposed to be completely absorbed by the water mass(Coffee mass assumed since it is modelled as liquid water) present and converted to its internal energy. This absorbed energy increases the specific internal energy (u) which in turn increases the temperature (T) of the water mass. It is given that u = 4.2 * T . That means temperature and internal energy (per unit mass ) has a linear relationship which means, whether we measure the change in temperature or change in total internal energy we can find the other from this relation. In the question the we know the total amount of heat energy radiated by oven per second, hence we can find the change in total internal energy (U) of water per second. Then we can also find out the change in temperature of whole of water per second using the previous relation. Then we can find out the time needed for increasing the temperature by 20⁰C.

Calculation:-

a) Rate of change of total internal energy(U) = dU/dt will be same as the rate of heat radiated by the oven, since we assume no heat loss, absorbed heat ll be used completely by water in increasing its internal energy at the same rate. That is dU/dt= 0.1 KW= 0.1 KJ / s.

Specific internal energy = u = internal energy in KJ / Kg of water

Total internal energy(U in KJ)= Specific internal energy(u in KJ/Kg) * Toatal mass of water(m)

=> dU = du * mass => dU/dt = du/dt *mass

=> du/dt = (dU/dt) / m = 0.1/0.2 = 0.5 KJ/Kg*S   (Ans)

b)

Hence u=4.2*T   => du/dt = 4.2 * dT/dt => 0.5= 4.2 * dT/dt => dT/dt = 0.5 / 4.2 = 0.12 K/S

=> dT = 0.12 dt

Let us say temp changes from T1 to T2 and time changes from t1 to t2

Now integrating the precious equation from T1 to T2 and t1 to t2

=> T2 - T2 = 0.12 * (t2 - t1)      Given that T2 - T1= 20⁰C = 20 K , => t2 - t1 = 20 / 0.12 = 166.66 seconds. (Ans)