A hot body cools from 90^ C to 70^ C in 4 minutes. In the next 6 minutes, its temperature drops from 70^ C to 50^ C . Assuming Newton's law of cooling holds true, the temperature of the surroundings is ________ ^ C .

Step-by-step solution

According to the average temperature approximation of Newton's law of cooling: T t = K ( T₁ + T₂ 2 - T₀ ) For the first interval (cooling from 90^ C to 70^ C in 4 minutes): 90 - 70 4 = K ( 90 + 70 2 - T₀ ) 5 = K(80 - T₀) ... (i) For the second interval (cooling from 70^ C to 50^ C in 6 minutes): 70 - 50 6 = K ( 70 + 50 2 - T₀ ) 10 3 = K(60 - T₀) ... (ii) Dividing equation (i) by equation (ii): 5 ( 10 3 ) = 80 - T₀ 60 - T₀ 15 10 = 80 - T₀ 60 - T₀ 1.5 = 80 - T₀ 60 - T₀ Cross-multiplying yields: 1.5(60 - T₀) = 80 - T₀