Answer: Given that:
Mass (m) = 20 g = 0.02 Kg,
temperature (T₁) = -10°C = -10+273 = 263 K
temperature (T₂) = 10°C = 10+273 = 283 K
Specific heat of water (Cp) = 4.187 KJ/Kg k
We know that Heat transfer (Q) = m. Cp.( T₂ - T₁)
= 0.02 × 4.187 × (283-263)
Q = 1.67 KJ
Heat transferred is 1.67 KJ
Step-by-step explanation: