1. Calculate the amount of heat (in calories) absorbed when 50.0g of water at 20

Question

1. Calculate the amount of heat (in calories) absorbed when 50.0g of water at 20degC spreads over your skin and warms to body temperature, 37degC.

2. A 102.5g sample of metal beads was heated in a water bath to 99.5degC. The metal was then added to a sample of water (25.7g) and the temperature of the water changed from an initial temperature of 20.0degC to a final temperature of 41.5degC. A)WHAT WAS THE TEMPERATURE OF THE METAL AT EQUILIBRIUM? B) CALCULATE THE SPECIFIC HEAT & ATOMIC MASS OF THE METAL. WHAT IS THE METAL?

Explanation / Answer

Heat given = dH = m*Cp*dt

dT = 37-20 = 17 C

Cp = 1 cal/C-g

m = 50g

dT = 50*17*1 = 850 Cal

ans 2)

Tf = 41.5 C

T1 = 20C...........initial water temp

T2 = 99.5 C...........initial metal temp

m1 = 25.7g

m2 = 102.5g

A> temperature of metal at equilibrium = 41.5 C

B> heat gained =heat lost

m1*Cw*(41.5-20) = m2*Cm*(99.5-41.5)

Cw = 4.186 J/g-C...........specific heat of water

so Cm = 0.389 J/g-C

(specific heat capacity) x (atomic mass) = 3R

by this relation

R = 8.314

atomic mass = 3*8.314/0.389 = 64.118 amu

this is close to copper so metal is copper

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