Question

Prelab: Exp 2 Calorimetry Name: Locker # 1. A student placed 25.421 g of metal in a coffee cup with exactly 50.0 grams of water. The initial temperature of the metal was 99.5 °C and the initial temperature of the water in the cup was 22.5 °C. If the final temp of the system was 31.8 °C what is the specific heat of the metal? 2. If the actual specific heat for the metal in question 1 is 1.35 J/g.°C, calculate the percent error. Percent error is calculated by the equation below accepted value-experimental value % error = x 100 accepted value

Answer 1

This is calorimetry problem,

in this kind of problem always use

heat loss= heat gain

m₁s1$\Delta T$1= m₂s_{2$\Delta T$2}

Griven -50.ogns = 22.5°02 mass of metal = 25.4219 1 mass of water Tempreture of u = T = 99.5 Pemberature of (T2) Final Temperature =T=31.8°c. 1 was by energy consen conservation Loss = Heat Heal -> 25.421XSX(99.5-31.8) = 50X40186 X (31,8 — 225) where s-specific Heat of metala 4.186 /gcc - specific heat of water hatter calculating we get S = 1.131 T/g °C. (D) .ersor - 11.55 – 10151 x 100 = 16.22% 1.35