Since the room temperature is 22 degrees Celsius,
[tex]\frac{dT}{dt}=k(T-22)[/tex]
Solving the differential equation,
[tex]\frac{1}{T-22} dT=k dt \\ \\ \int \frac{1}{T-22} dT=\int k dt \\ \\ \ln|T-22|=kt+C[/tex]
When t=0, the temperature is 100 degrees.
[tex]\ln 78=C \\ \\ \implies \ln|T-22|=kt+\ln 78[/tex]
When t=1, the temperature is 80 degrees.
[tex]\ln 58=k+\ln 78 \implies k=\ln(29/39) \\ \\ \implies \ln|T-22|=t \ln(29/39)+\ln 78[/tex]
Setting t=4,
[tex]\ln|T-22|=4\ln(29/39)+\ln 78 \\ \\ T=22+e^{4\ln(29/39)+\ln 78} \approx \boxed{46}[/tex]
