How much heat is absorbed when a block of copper of mass 0.05kg and specific heat capacity 390Jkg^-1K^-1 is heated from 20^oC to 70^oC?

39 . 98 x 10^-1J

9.75 x 10²J

3.98 x 10³J

9.75 x 10³J

Correct Option

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3 years ago

Q = CMΔθ

where C = 390Jkg-1K-1
M = 0.05 kg
Δθ = 70 - 20 = 50

so therefore,
Q = 390 * 0.05 * 50
Q = 975J
in standard form, Q = 9.75 * 10^2

3 years ago

My school check Dis mistake and correct it

2 years ago

ceoofwahala

20th June, 2026

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20th June, 2026

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21st May, 2026

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How much heat is absorbed when a block of copper of mass 0.05kg and specific heat capacity 390Jkg-1K-1 is heated from 20oC to 70oC?