The amount of heat required to convert 1 gm of ice at -10° C to steam at 100° C is :

heat required $\text{Q}={\text{Q}}_1{\text{+Q}}_2{\text{+Q}}_3{\text{+Q}}_4$

$\begin{array}{rcl}total amount of heat required is Q& =& Q_1+Q_2+Q_3+Q_4\\ here Q_1& =& amount of heat required to convert ice at -{10}^{0}C to ice at 0^{0}C=m{C}_1∆T_1\\ Q_2& =& amount of heat required to convert ice at 0^{0}C to water at 0^{0}C =m{L}_1\\ Q_3& =& amount of heat required to convert water at 0^{0}C to water at {100}^{0}C=m{C}_2∆T_2\\ Q_4& =& amount of heat required to convert water at {100}^{0}C to steam at {100}^{0}C=m{L}_2\\ here m is mass & =& 1gm\\ C_1& =& specific heat of ice=0.5calorie/g/{}^{0}c\\ & & {\text{ΔT}}_{\text{1}}\text{=temperature change from -10 to 0=0-(-10)=10°C}\\ L_{\text{1}}& =& latent heat of fusion of ice=80 calorie/g\\ C_2& =& specific heat of water=1calorie/g/{}^{0}c\\ & & {\text{ΔT}}_2\text{=temperature change from 0 to 100=100-0=100°C}\\ L_2& =& latent heat of vaporisation=540 calorie/g\\ & & substitute above values in equation of Q\\ & & {\text{Q=mC}}_{\text{1}}{\text{ΔT}}_{\text{1}}{\text{+mL}}_{\text{1}}{\text{+mC}}_{\text{2}}{\text{ΔT}}_{\text{2}}{\text{+mL}}_{\text{2}}\\ Q& =& 1\times 0.5\times 10+1\times 80+1\times 1\times 100+1\times 540\\ Q& =& 5+80+100+540=725 cal.\end{array}$