It takes 210 cal to raise the temperature of 340 g of lead from 0 to 10 C. What is the specific heat of lead?
2007-12-06 16:32:44 · 1 answers · asked by jim m 3 in Science & Mathematics ➔ Physics
i need to answer in kcal/(kg K)
2007-12-06 17:05:13 · update #1
Q = m c ΔT
Q = heat gained or lost
m = mass
c = specific heat
ΔT = temperature change
Plug in your values and solve for c; units will be cals / g-°C
2007-12-06 16:49:59 · answer #1 · answered by jgoulden 7 · 0⤊ 0⤋
We will first convert the given values to the required units: Q = 210 cal = 0.210 kcal m = 340 g = 0.340 kg T1 = 0°C = 273.15 K T2 = 20°C = 293.15 K Then solve for C (Specific heat) using the formula Q = mC∆T C = Q / (m∆T) ∆T = T2 - T1 = 293.15 K - 273.15 K = 20.0 K C = 0.210 kcal / (0.340 kg * 20.0 K) = 0.0309 kcal/kg-K Hope that helps! XD
2016-05-21 23:17:50 · answer #2 · answered by ? 3 · 0⤊ 0⤋