A sample of brass has a specific heat of 380 J/kg°C. If 1410 Joules of heat is applied to a 1.5 kg piece and the initial temperature is 2.50°C, what will be the final temperature? Q = (m) (Csp) (ΔT) ΔT = Tfinal - Tinitial

a

.98°C

b

.003°C

c

5.5°C

d

3°C

Question

Chemistry

Adrianna Decker

30 July, 08:36

A sample of brass has a specific heat of 380 J/kg°C. If 1410 Joules of heat is applied to a 1.5 kg piece and the initial temperature is 2.50°C, what will be the final temperature? Q = (m) (Csp) (ΔT) ΔT = Tfinal - Tinitial

a

.98°C

b

.003°C

c

5.5°C

d

3°C

+2

Answers (1)

Know the Answer?

Yazmin Franklin · Answer 1

Heat energy supplied raises the temperature of a mass of material. the formula is as follows

Q = m x Csp x ΔT

Q - heat energy supplied

m - mass of brass

Csp - specific heat

ΔT - change in temperature

substituting the values in the equation

1410 J = 1.5 kg x 380 J/kg°C x ΔT

ΔT = 2.47 °C

since ΔT = T final - T initial

and T initial = 2.50 °C

2.47 = x - 2.50

x = 4.97 °C

answer should be 4.97 °C

A sample of brass has a specific heat of 380 J/kg°C. If 1410 Joules of heat is applied to a 1.5 kg piece and the initial temperature is 2.50°C, what will be the final temperature? Q = (m) (Csp) (ΔT) ΔT = Tfinal - Tinitiala.98°Cb.003°Cc5.5°Cd3°C

A sample of brass has a specific heat of 380 J/kg°C. If 1410 Joules of heat is applied to a 1.5 kg piece and the initial temperature is 2.50°C, what will be the final temperature? Q = (m) (Csp) (ΔT) ΔT = Tfinal - Tinitial

a

.98°C

b

.003°C

c

5.5°C

d

3°C