Given that the specific heat of water is 4.18 kj/(kg·°c), how much heat does it take to raise the temperature of 3.5 kg of water from 25°c to 55°c?

The heat required =438900 J

Explanation:

mass= 3.5 Kg

change in temperature=ΔT=55-25=30°C

specific heat of iron =4.18 kJ/(kg·°C)=4180 J/(kg·°C)

the formula for heat is given by

Q= m C ΔT

Q=3.5 (4180) (30)

Q=438900 J

the assestment will probley apele with cause and effect

dec 23, 2016 - a load of bricks weighing 680 n a distance of 10 m across a parking lot. if if the worker pushes the cart with a constant force of 209 n, what work= force x distance = 220 n x 10 m = 2200 n.m = 2200 j i don't know where the person got it from, but the initial part, work= force x distance me.