Note: The question is incomplete. The complete question is given below : Suppose a substance has a heat of fusion equal to 45 cal/g and a specific heat of 0.75 cal/g°C in the liquid state. If 5.0 kcal of heat are applied to a 50 g sample of the substance at a temperature of 24°C, what will its new temperate be? What state will the sample be in? (melting point of the substance = 27°C; specific heat of the solid =0.48 cal/g°C; boiling point of the substance = 700°C) Explanation: 1.a) Heat energy required to raise the temperature of the substance to its melting point, H = mcΔT Mass of solid sample = 50 g; specific heat of solid = 0.75 cal/g; ΔT = 27 - 24 = 3 °C H = 50 × 0.75 × 3 = 112.5 calories b) Heat energy required to convert the solid to liquid at its melting point at 27°C, H = m×l, where l = 45 cal/g H = 50 × 45 = 2250 cal c) Total energy used so far = 112.5 cal + 2250 cal = 2362.5 calories. Amount of energy left = 5000 - 2362.5 = 2637.5 cal The remaining energy is used to heat the liquid H = mcΔT Where specific heat of the liquid, c = 0.75 cal/g/°C, H = 2637.5 cal, ΔT = temperature change 2637.5 = 50 × 0.75 x ΔT ΔT = 2637.5 / ( 50*0.75) ΔT = 70.3 °C Final temperature of sample = (70.3 + 27) °C = 97.3 °C The substance will be in liquid state at a temperature of 97.3 °C
