The inner and outer surfaces of a furnace wall, 25 cm thick, are at 300 degree Celsius and 30 degree Celsius. Here thermal conductivity is given by the relationK = (1.45 + 0.5 * 10-5 t2) KJ/m hr degWhere, t is the temperature in degree centigrade. Calculate the heat loss per square meter of the wall surface area?

• A. 1355.3 kJ/m2 hr
• B. 2345.8 kJ/m2 hr
• C. 1745.8 kJ/m2 hr
• D. 7895.9 kJ/m2 hr