To what temperature in degrees F would 26 lbm of a material

To what temperature (in degrees F) would 26 lb_m of a material at 25C be raised if 131 Btu of heat is supplied? Assume that the c_p value for this material is 408 J/kg-K.

Solution

Given

mass, m = 26 lbm = 26 X 0.454 = 11.804 kg

Initial temperature, T_i = 25^oC

Heat supplied, Q = 131 Btu = 131 X 1055.06 = 138212.86 J

C_p = 408 J/kgK

Q = mC_p(T_f - T_i)

where T_f is the final temperature

138212.86 = 11.804 X 408 X (T_f - 25)

T_f = 53.7^oC...................................................(Answer)